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Structural
mechanics
Charles Ezra
Greene, Albert
Emerson Greene
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WORKS OF C. E. GREENE
PUBLISHED BY
JOHN WILEY & SONS.
Graphics for Engineers, Architects, and Builders.
A manual for designers, and a text-book for scientific
schools.
Trasses and Arches:
Analyzed and Discussed by Graphical Methods. In
three parts— published separately.
Part I. Roof Trusses :
Diagrams for Steady Load, Snow and Wind. 8vo, 80
pp., 3 folding plates. Revised Edition. $1.35.
Pert II. Bridge Trusses:
Single, Continuous, and Draw Spans; Single and Mul-
tiple Systems; Straight and Inclined Chords. 8vo, 190
pp., 10 folding plates. Fifth Edition, Revised. $3.50.
Part III. Arches,
In Wood, Iron, and Stone, for Roofs, Bridges, and V all-
openings ; Arched Ribs and Braced Arches; Stresses
from Wind and Change of Temperature. 8vo, 194 pp.,
8 folding plates. Third Edition, Revised. $2.50.
Structural Mechanics.
Comprising the Strengtn and Resistance of Materials
and Elements of Structural Design. With Examples
and Problems. By the late Charles E. Greene, A.M.,
C.E. New Edition, Revised and Enlarged by A. E.
Greene. 8vo, viii 4- 244 pages, 99 figures. $3.50 nrt.
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STRUCTURAL MECHANICS
COMPRISING THE
STRENGTH AND RESISTANCE OF MATERIALS AND
ELEMENTS OF STRUCTURAL DESIGN
WITH EXAMPLES AND PROBLEMS
BY
CHARLES E. GREENE, A.M., C.E.
LATE PROFESSOR OF CIVIL ENGINEERING, UNIVERSITY OF MICHIGAN
REVISED BY A. E. GREENE
ASSISTANT PROFESSOR OF CIVIL ENGINEERING, UNIVERSITY OF MICHIGAN
SECOND EDITION
FIRST THOUSAND
NEW YORK
JOHN WILEY & SONS
London: CHAPMAN & HALL, Limited
1905
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Copyright, 1897,
BY
Charles E. Greene
Copyright, 1905,
BY
A. E. Greene
ROBERT DRUMMOND, PRINTER, NEW YORK
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94192
•Ga^
PREFACE.
The author, in teaching for many years the subjects embraced
in the following pages, has found it advantageous to take at
first but a portion of what is included in the several chapters,
and, after a general survey of the field, to return and extend the
investigation more in detail. Some of the sections, therefore,
are not leaded and can be omitted at first reading. A few of
the special investigations may become of interest only when the
problems to which they relate occur in actual practice.
It is hoped that this book will be serviceable after the class-
room work is concluded, and reference is facilitated by a more
compact arrangement of the several matters than the course
suggested above would give. The attempt' has been made to
deal with practicable cases, and the examples for the most part
are shaped with that end in view. A full index will enable one
to find any desired topic.
The treatment of the subject of internal stress is largely
graphical. All the constructions are simple, and the results,
besides being useful in themselves, shed much light on var.'ous
problems. The time devoted to a careful study of the chapter
in question will be well expended.
The notation is practically uniform throughout the book,
and is that used by several standard authors. Forces and mo-
ments are expressed by capital letters, and unit loads and stresses
by small letters. The coordinate x is measured along the length
of a piece, the coordinate y in the direction of variation of stress
iii
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IV PREFACE.
in a section, and z is the line of no variation of stress, that is,
the line parallel to the moment axis.
One who has mastered the subjects discussed here can use
the current formulas, the pocket-book rules, and tables, not
blindly, but with discrimination, and ought to be prepared to
design intelligently.
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TABLE OF CONTENTS.
PAGB
Introduction i
CHAPTER I.
Action of a Piece under Direct Force. 6 — n
CHAPTER H.
Materials 19 - c
CHAPTER III.
Beams 38 -
CHAPTER IV.
Moments op Inertia of Plane Areas 71 i>
CHAPTER V.
Torsion 81
CHAPTER VI.
Flexure and Deflection of Simple Beams 87 - ' ^
CHAPTER VII.
Restrained Beams: Continuous Beams 107
CHAPTER VIII.
Pieces under Tension 127 "
CHAPTER IX.
Compression Pieces: Columns, Posts, and Struts 137 - L *
CHAPTER X.
Safe Working Stresses 153 - c
T
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Vl TABLE OF CONTENTS.
CHAPTER XI.
PAGB
Internal Stress: Change of Form 167 - G>
CHAPTER XII.
Rivets: Pins 192 ^ V
CHAPTER XIII.
Envelopes: Boilers. Pipes, Dome 203 *b
CHAPTER XIV.
Plate Girder. 221 !
CHAPTER XV.
Springs: Plates 229 - '
CHAPTER XVI.
Reinforced Concrete. 235 c
/r
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NOTATION.
6, breadth of rectangular beam.
C, shearing modulus of elasticity.
d, diameter.
E, modulus of elasticity, Young's modulus.
F, shear in beam.
/, unit stress.
h, height of rectangular beam.
J, rectangular moment of inertia.
i, slope of elastic curve.
/, polar moment of inertia.
k, a numerical coefficient.
/, length of member.
A, unit change of length.
M, bending or resisting moment.
P, reaction of beam; load on tie or post.
p, unit stress; unit pressure in envelope, Ch. Xill.
q, unit shear.
R, radius of circle.
r, radius of gyration; radius of envelope, Ch. Xlll.
p, radius of curvature.
5, area of cross-section.
T, torsional moment ; stress in envelope, Ch. XIII.
v y deflection of beam.
W, concentrated load on beam.
w, intensity of distributed load on beam.
yu distance from neutral axis to extreme fibre of beam.
x, y, z, coordinates of length, depth, and breadth of beam.
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STRUCTURAL MECHANICS.
INTRODUCTION.
i. External Forces. — The engineer, in designing a new
structure, or critically examining one already built, determines
from the conditions of the case the actual or probable external
forces which the structure is callejl upon to resist. He may
then prepare, either by mathematical calculations or by graphical
methods, a sheet which shows the maximum and minimum
direct forces of tension and compression which the several pieces
or parts of the structure are liable to experience, as well as the
bending moments on such parts as are subjected to them.
These forces and moments are determined from the require-
ments of equilibrium, if the pieces are at rest. For forces acting
in one plane, a condition which suffices for the analysis of most
cases, it is necessary that, for the structure as a whole, as well
as for each piece, there shall be no tendency to move up or down,
to move to the right or left, or to rotate. These limitations
are usually expressed in Mechanics as, that the sum of the X
forces, the sum of the Y forces, and the sum of the moments
shall each equal zero.
If the structure is a machine, the forces and moments in
action at any time, and their respective magnitudes, call for a
consideration of the question of acceleration or retardation of
the several parts and the additional maximum forces and
moments called into action by the greatest rate of change of
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2 STRUCTURAL MECHANICS.
motion at any instant. Hence the weight or mass of the mov-
ing part or parts is necessarily taken into account.
Finally, noting the rapidity and frequency of the change
of force and moment at any section of any piece or connection,
the engineer selects, as judgment dictates, the allowable stresses
of the several kinds per square inch, making allowance for the
effect of impact, shock, and vibration in intensifying their action,
and proceeds to find the necessary cross-sections of the parts
and the proportions of the connections between them. As all
structures are intended to endure the forces and vicissitudes to
which they are usually exposed, the allowable unit stresses,
expressed in pounds per square inch, must be safe stresses.
It is largely with the development of the latter part of this
subject, after the forces have been found to which the several
parts are liable, that this book is concerned.
2. Ties, Struts, and Beams. — There are, in general, three
kinds of pieces in a frame or structure: ties or tension members;
columns, posts, and struts or compression members; and beams,
which support a transverse load and are subject to bending and
its accompanying shear. A given piece may also be, at the
same time, a tie and a beam, or a strut and a beam, and at dif-
ferent times a tie and a strut.
3. Relation of External Forces to Internal Stresses. — The
forces and moments which a member is called upon to resist,
and which may properly be considered as external to that mem-
ber, give rise to actions between all the particles of material of
which such a member is composed, tending to move adjacent
particles from, towards, or by one another, and causing change
of form. There result internal stresses, or resistances to dis-
placement, between the several particles.
These internal stresses, or briefly stresses, must be of such
kind, magnitude, distribution, and direction, at any imaginary
section of a piece or structure, that their resultant force and
moment will satisfy the requirements of equilibrium or change
of motion with the external resultant force and moment at that
section; and no stress per square inch can, for a correct design,
be greater than the material will safely bear. Hence may be
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INTRODUCTION. 3
determined the necessary area and form of the cross-section at
the critical points, when the resultant forces and moments are
known.
4. Internal Stress. — There are three kinds of stress, or action
of adjacent particles one on the other, to which the particles of
a body may be subjected, when external forces and its own weight
are considered, viz. : tensile stress, tending to remove one particle
farther from its neighbor, and manifested by an accompanying
stretch or elongation of the body; compressive stress, tending to
make a particle approach its neighbor, and manifested by an
accompanying shortening or compression of the body; and
shearing stress, tending to make a particle move or slide laterally
with reference to an adjacent particle, and manifested by an
accompanying distortion. Whether the stress produces change
of form, or the attempted change of form gives rise to internal
stresses as resistances, is of little consequence; the stress between
two particles and the change of position of the particles are
always associated, and one being given the other must exist.
5. Tension and Shear, or Compression and Shear. — If the
direction of the stress is oblique, that is, not normal or perpen-
dicular, on any section of a body, the stress may be resolved into
a tensile or compressive stress normal to that section, and a
tangential stress along the section, which, from its tendency to
cause sliding of one portion of the body by or along the section,
has been given the name of shear, from the resemblance to the
action of a pair of shears, one blade passing by the other along
the opposite sides of the plane of section. Draw two oblique
and directly opposed arrows, one on either side of a straight line
representing the trace of a sectional plane, decompose those
oblique stresses normally and tangentially to the plane, and
notice the resulting directly opposed tension or compression,
and the shear. Hence tension and shear, or compression and
shear, may be found on any given plane in a body, but tension
and compression cannot simultaneously occur at one point in
a given area.
6. Sign of Stress.— Ties are usually slender members; struts
have larger lateral dimensions. Longitudinal tension tends to
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4 STRUCTURAL MECHANICS.
diminish the cross-section of the piece which carries it, and hence
may conveniently be represented by — , the negative sign; longi-
tudinal compression tends to increase the cross-sectional area
and may be called + or positive. Shear, being at right angles
to the tension and compression in the preceding illustration, has
no sign; and lies, in significance, between tension and com-
pression. If a rectangular plate is pulled in the direction of two
of its opposite sides and compressed in the direction of its other
two sides, there will be some shearing stress on every plane of
section except those parallel to the sides, and nothing but shear
on two certain oblique planes, as will be seen later.
7. Unit Stresses. — These internal stresses are measured by
units of pounds and inches by English and American engineers,
and are stated as so many pounds of tension, compression, or
shear per square inch, called unit tension, compression, or shear.
Thus, in a bar of four square inches cross-section, under a total
pull of 36,000 pounds centrally applied, the internal unit tension
is 9,000 pounds per square inch, provided the pull is uniformly
distributed on the particles adjacent to any cross-section. If
the pull is not central or the stress not uniformly distributed,
the average or mean unit tensile stress is still 9,000 pounds.
If an oblique section of the same bar is made, the total force
acting on the particles adjacent to the section is the same as before,
but the area of section is increased; hence the unit stress, found
by dividing the force by the new area, is diminished. The stress
will also be oblique to the section, as its direction must be that
of the force. When the unit stress is not normal to the plane
of section on which it acts, it can be decomposed into a normal
unit tension and a unit shear. See § 151.
When the stress varies in magnitude from point to point,
its amount on any very small area (the infinitesimal area of the
Calculus) may be divided by that area, and the quotient will
be the unit stress, or the amount which would exist on a square
inch, if a square inch had the same stress all over it as the very
small area has.
8. Unit Stresses on Different Planes not to be Treated as
Forces. — It will be seen, upon inspection of the results of analyses
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INTRODUCTION. 5
which come later, that unit stresses acting on different planes
must not be compounded and resolved as if they were forces.
But the entire stress upon a certain area, found by multiplying
the unit stress by that area, is a force, and this force may be
compounded with other forces or resolved, and the new force
may then be divided by the new area of action, and a new unit
stress be thus found.
Some persons may be assisted in understanding the analysis
of problems by representing in a sketch, or mentally, the unit
stresses at different parts of a cross-section by ordinates which
make up, in their assemblage, a volume. This volume, whose
base is the cross-section, will represent or be proportional to
the total force on the section. The position of the resultant
force or forces, i.e., traversing the centre of gravity of the volume,
the direction and law of distribution of the stress are then quite
apparent.
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CHAPTER I.
ACTION OF A PIECE UNDER DIRECT FORCE.
9. Change of Length under an Applied Force. — Let a uniform
bar of steel have a moderate amount of tension applied to its
two ends. It will be found, upon measurement, to have increased
in length uniformly throughout the measured distance. Upon
release of the tension the stretch disappears, the bar resuming
its original length. A second application of the same amount of
tension will cause the same elongation, and its removal will be
followed by the same contraction to the original length. The
bar acts like a spring. This elastic elongation (or shortening
under compression) is manifested by all substances which have
definite form and are used in construction; and it is the cause
of such changes of shape as structures, commonly considered
rigid, experience under changing loads. The product of the
elongation (or shortening) into the mean force that produced it
is a measure of the work done in causing the change of length.
As the energy of a moving body can be overcome only by work
done, the above product becomes of practical interest in structures
where moving loads, shocks, and vibrations play an important
part.
10. Modulus of Elasticity. — If the bar of steel is stretched
with a greater force, but still a moderate one, it is found by care-
ful measurement that the elongation has increased with the
force; and the relationship may be laid down that the elongation
per linear inch is directly proportional to the unit stress on the
cross-section per square inch.
The ratio of the unit stress to the elongation per unit of length
is denoted by E, which is termed the modulus 0} elasticity of the
6
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ACTION OF A PIECE UNDER DIRECT FORCE. 7
material, and is based, in English and American books, upon
the pound and inch as units. If P is the total tension in pounds
applied to the cross-section, 5, measured in square inches, M
the elongation in inches, produced by the tension, in the pre-
viously measured length of / inches, and / the stress per square
inch of cross-section,
*-*-/. i-L
Hence, if E has been determined for a given material, the
stretch of a given bar under a given unit stress is easily found.
Since the elongation per unit of length, ^, is merely a ratio
and is the same whatever system of units is employed, E will be
expressed in the same units as /.
Example. — A bar of 6 sq. in. section stretches 0.085 in. in a measured
length of 120 in. under a pull of 120,000 lb.
„ 120,000X120
"~6Xoo87~" = 28,200,000 lb. per sq. in.
If the stress were compressive, a similar modulus would
result, which will be shown presently to agree with the one just
derived.
If one particle is displaced laterally with regard to its neighbor,
under the action of a shearing stress, a modulus of shearing elas-
ticity will be obtained, denoted by C, the ratio of the unit shear
to the angle of distortion. See § 173.
11. Stress-stretch Diagram. — The elongations caused in a
certain bar, or the stretch per unit of length, may be plotted
as abscissas, and the corresponding forces producing the stretch,
or the unit stresses per square inch, may be used as ordinates,
defining a certain curve, as represented in Fig. 1. This curve
can be drawn on paper by the specimen itself, when in the testing-
machine, if the paper is moved in one direction to correspond
with the movement of the poise on the weighing arm, and the
pencil is moved at right angles by the stretch of the specimen.
A similar diagram can be made for a compression specimen,
and may be drawn in the diagonally opposite quadrant. Pull
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8 STRUCTURAL MECHANICS.
will then be rightly represented as of opposite sign to thrust,
and extension will be laid off in the opposite direction to shorten-
ing or compression.
12. Work of Elongation. — If the different unit stresses
applied to the bar are laid off on O Y as ordinates and the result-
ing stretches per unit of length on O X as abscissas, the portion
of the diagram near the origin will be found to be a straight line,
more or less oblique, according to the scale by which the elonga-
tions are platted. The elongation varies directly as the unit
stress, beginning with zero. Hence the mean force is £P, and
the work done in stretching a given bar with a given force, if
the limit of elastic stretch is not exceeded, is
P F*l
Work = — >l/=-
2 2ES'
It may be seen that the work done in stretching the bar is
represented by the area included between the base line or axis,
the curve O A, and the ordinate at A. It also appears that E
may be looked upon as the tangent of the angle X O A. A
material of greater resistance to elongation will give an angle
greater than X O A and vice versa.
Example. — A bar 20 ft. = 240 in. long and 3 sq. in. in section is
to have a stress applied of 10,000 lb. per sq. in.; if £=28,000,000,
the work done on the bar will be
30^- 3 o,ooo-a4o -^
2-28,000,000-3
and the stretch will be 1,286-5-15,000=0.086 in.
13. Permanent Set. — While the unit stress may be grad-
ually increased with corresponding increase of stretch, and
apparently complete recovery of original length when the bar
is released, there comes a time when very minute and delicate
measurements show that the elongation has increased in a slightly
greater degree than has the stress. The line O A at and beyond
such a point must therefore be a curve, concave to the axis of X.
If the piece is now relieved from stress, it will be found that the
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ACTION OF A PIECE UNDER DIRECT FORCE. 9
bar has become permanently lengthened. The amount of this
increase of length after removal of stress is called set, or perma-
nent set, and the unit stress for which a permanent set can first
be detected is known as the elastic limit. As the elongation
itself is an exceedingly small quantity, even when measured in
a length of many inches, and the permanent set is, in the begin-
ning, a quantity far smaller and hence more difficult of deter-
mination, the place where the straight line OA first begins to
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vV" 1
^
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c*y
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-<?
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4
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.
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» X
Fig. 1
curve is naturally hard to locate, and the accurate elastic limit
is therefore uncertain. Some contend that O A itself is a curve
of extreme flatness. The common or commercial elastic limit lies
much farther up the curve, where the permanent set becomes
decidedly notable.
If, after a certain amount of permanent set has occurred
in a bar, and the force which caused it has been removed, a
somewhat smaller force is repeatedly applied to the bar, the
piece will elongate and contract elastically to the new length,
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10 STRUCTURAL MECHANICS.
i.e., old length plus permanent set, just as if the unit stress were
below the elastic limit.
14. Yield-point. — The unit stress increasing, the elongation
increases and the permanent set increases until a unit stress B
is reached, known as the yield-point (or commercial elastic limit,
or common elastic limit), which causes the bar to yield or draw
out without increase of force, and, as the section must decrease,
apparendy with decreasing power of resistance. There will
then be a break of continuity in the graphic curve. A decided
permanent elongation of the bar takes place at this time — suf-
ficient to dislodge the scale from the surface of a steel bar, if
left as it comes from the rolls or hammer. The weighing beam
of the testing-machine falls, from the diminished resistance
just referred to, and remains stationary while the bar is elongating
for a sensible interval of time. Hence, for steel, the yield-point,
or common elastic limit, is easily determined by what is known
as the "drop of the beam." The remainder of the curve, up to
the breaking-point, is shown in the figure.
15. Elastic Limit Raised. — For stresses above the yield-
point also, a second application and release of stress will give
an elastic elongation and contraction as before the occurrence
of set, as shown by line E F, so that a new elastic limit may be
said to be established. The stretch due to any given stress may
be considered to be the elastic elongation plus the permanent
set; and, for repetitions of lesser forces, the bar will give a line
parallel to O A, as if drawn from a new origin on O X, distant
from O the amount of the permanent set,
If the line O A is prolonged upwards, it will divide each
abscissa into two parts, of which that on the left of OA will
be the elastic stretch, and that on the right of O A the permanent
set for a given unit stress.
16. Work of Elongation, for Stress above Yield-point. — The
area below the curve, and limited by any ordinate G D, will
be the work done in stretching the bar with a force represented
by the product of that ordinate into the bar's cross-section, and
if a line be drawn from the upper end of that ordinate parallel
to O A, the triangle C D G will give the work done in elastic
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ACTION OF A PIECE UNDER DIRECT FORCE. ir
stretch and the quasi-parallelogram O B D C will show the
permanent work of deformation done on the bar. It should
be remembered that, as the bar stretches, the section decreases,
and that the unit stress cannot therefore be strictly represented
by P-5-5, if S is the original cross-section. The error is not of
practical consequence for this discussion.
17. Ultimate or Breaking Strength. — If the force applied in
tension to the bar is increased, a point will next be reached where
a repeated application of the same force causes a successive increase
in the permanent elongation. As this phenomenon means a
gradual drawing out, final failure by pulling asunder is only
a matter of a greater or less number of applications of the force.
While the bar is apparently breaking under this force, the rapid
diminution of cross-section near the breaking point actually gives
a constantly rising unit stress, as is seen by the dotted curve of
the figure.
If, however, the force is increased without pause from the
beginning, the breaking force will be higher, as might be ex-
pected; since much work of deformation is done upon the bar
before fracture. The bar would have broken under a somewhat
smaller force, applied statically for a considerable time.
The elongation of the bar was uniform per unit of length
during the earlier part of the test. There comes a time when a
portion or section of the bar, from some local cause, begins to
yield more rapidly than the rest. At once the unit stress at
that section becomes greater than in the rest of the bar, by reason
of decrease of cross-section, and the drawing out becomes in-
tensified, with the result of a great local elongation and necking
of the specimen and an assured final fracture at that place. If
the bar were perfectly homogeneous, and the stress un'formly
distributed, the bar ought to break at the middle of the length,
where the flow of the metal is most free.
It is customary to determine, and to require by specification,
in addition to elastic limit and ultimate strength (on one con-
tinuous application of increasing load), the per cent, of elonga-
tion after fracture (which is strictly the permanent set) in a cer-
tain original measured length, usually eight inches, and the per
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12 STRUCTURAL MECHANICS.
cent, of reduction of the original area, after fracture, at the point
of fracture. As the measured length must include the much
contracted neck, the average per cent, of elongation is given
under these conditions. A few inches excluding the neck would
show less extension, and an inch or two at the neck would give
a far higher per cent, of elongation. The area between the axis
of X, the extreme ordinate, and the curve will be the work of
fracture, if S is considered constant, and will be a measure of
the ability of the material to resist shocks, blows, and vibrations
before fracture. It is indicative of the toughness or ductility of
the material.
The actual curve described by the autographic attachment
to a testing-machine is represented by the full line; the real
relation of stress per square inch to the elongation produced,
when account is taken of the progressive reduction of sectional
area, is shown by the dotted line. The yield-point, or common
elastic limit, is very marked, there appearing to be a decided
giving way or rearrangement of the particles at that value of
stress. The true elastic limit is much below that point.
18. Effect of a Varying Cross-section. — If a test specimen
is reduced to a smaller cross-section, by cutting out a curved
surface, for only a short distance as compared with its transverse
dimensions, it will show a greater unit breaking stress, as the
metal does not flow freely, and lateral contraction of area is
hindered. But, if the portion of reduced cross-section joins the
rest of the bar by a shoulder, the apparent strength is reduced,
owing to a concentration of stress on the particles at the corner
as the unit stress suddenly changes from the smaller value on the
larger section to the greater unit stress on the smaller cross-
section.
19. Compression Curve. — A piece subjected to compression
will shorten, the particles being forced nearer together, and the
cross-section will increase. It might be expected, and is found
by~ experiment to be the case, that, in the beginning, the resistance
of the particles to approach would be like their resistance to
separation under tension, so that the tension diagram might be
prolonged through the origin into the third quadrant, reversing
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ACTION OF A PIECE UNDER DIRECT FORCE. 13
the sign of the ordinate which represents unit stress and of the
abscissa which shows the corresponding change of length. As
this part of the diagram is a straight line, it follows that the value
of £, the elastic modulus for compression, is the same as that
for tension. After passing the yield-point the phenomena of com-
pression are not so readily determined, as fracture or failure by
compressive stress is not a simple matter, and the increase of
sectional area in a short column of ductile material will interfere
with the experiment. In long columns and with materials not
ductile, failure takes place in other ways, as will be explained
later.
The compression curve is here shown in the same quadrant
with the tension curve for convenience and comparison.
20. Resilience. — By definition, § 10, if / is the unit stress
per square inch and il the stretch of a bar of length /, in inches,
the modulus of elasticity E=f + X, provided / does not exceed
the elastic limit. Also the work done in stretching a bar inside
the elastic limit, by a force P, gradually applied, that is, beginning
with zero and increasing with the stretch, is the product of the
mean force, £P, into the stretch, or
Work done = ^P-U^— •%=-•¥ -SI.
1 2 E 2 E
The amount of work which must be done upon a piece in order
to produce the safe unit stress, /, in it is the resilience of the piece.
SI is the volume of the bar; P+E is called the modulus of re-
silience, when / is the elastic limit, or sometimes the maximum
safe unit stress. This modulus depends upon the quality of the
material, and, as it is directly proportional to the amount of
work that can safely be done upon the bar by a load, it is a
measure of the capacity of a certain material for resisting or
absorbing shock and impact without damage. For a particular
piece, the volume SI is also a factor as above. A light structure
will suffer more from sudden or rapid loading than will a
heavier one of the same material, if proportioned for the same
unit stress.
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14 STRUCTURAL MECHANICS.
21. Work Done Beyond the Elastic Limit. — The work done
in stretching a bar to any extent is, in Fig. i, the area in the
diagram between the curve from the origin up to any point,
the ordinate to that point, and the axis of abscissas, provided
the ordinate represents P, and the abscissa the total stretch.
Further, it may be seen from the figure that, if a load applied
to the bar has exceeded the yield-point, the bar, in afterwards
contracting, follows the line F E; and, upon a second applica-
tion of the load, the right triangle of which this line is the hypothe-
nuse will be the work done in the second application, a smaller
quantity than for the first application. But, if the load, in its
second and subsequent applications, possesses a certain amount
of energy, by reason of not being gently or slowly applied, this
energy may exceed the area of the triangle last referred to, with
the result that the stress on the particles of the bar may become
greater than on the first application. Indeed it is conceivable
that this load may be applied in such a way that the resulting
unit stress may mount higher and higher with repeated applica-
tions of load, until the bar is broken with an apparent unit stress
P-f-5, far less than the ultimate strength, and one which at first
was not much above the yield-point. If the load in its first
application is above the yield-point of the material, and it is
repeated continuously, rupture will finally occur.
What is true for tensile stresses is equally true for compressive
stresses, except that the ultimate strength of ductile materials
under compression is uncertain and rather indefinite.
22. Sudden Application of Load. — If a steel rod, io feet =
120 inches long, and one square inch in section, with E =
28,000,000, is subjected to a force increasing gradually from o
to 12,000 lb. longitudinal tension, its stretch will be 1 2,000 X
120-^28,000,000=0.05 in., and the work done in stretching the
bar will be J X 12,000X0.05 =300 in.-lb.
But if the 12,000 lb. is suddenly applied, as by the extremely
rapid loading of a structure of which the rod forms a part, or
by the quick removal of a support which held this weight at the
lower end of the rod, the energy due to a fall of 0.05 in. is 12,000 X
0.05 = 600 in.-lb., while the work done upon the rod is but 300 in.-lb.
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ACTION OF A PIECE UNDER DIRECT FORCE. 15
as before. The excess of 300 in.-lb. of energy must be absorbed
by the rod and it will continue to stretch until the energy due to
the fall equals the work done upon the rod or until it has stretched
0.10 in. The stress in the rod is then 24,000 lb. or twice the
suddenly applied load; the energy due to the fall is 12,000X0.10
and the work done upon the rod is JX 24,000X0.10 = 1,200 in. -lb.
As equilibrium does not exist between the external force and the
internal stress, the rod will contract and then undergo a series of
longitudinal vibrations of decreasing amplitudes, finally settling
down to a stretch of 0.05 in., when the extra work of accelera-
tion has been absorbed. The work of acceleration on the mass
of the bar is neglected.
A load applied to a piece with absolute suddenness produces
twice the deformation and twice the stress which the same load
does if applied gradually. Stresses produced by moving loads
on a structure are intermediate in effect between these two extremes^
depending upon rapidity or suddenness of loading. Hence it is
seen why the practice has arisen of limiting stresses due to moving
loads apparently to only one-half of the values permitted for
those caused by static loading.
For resilience or work done in deflection of beams, see § 100.
23. Granular Substances tinder Compression. — Failure by
Shearing on Oblique Planes. Blocks of material, such as cast
iron, sandstone, or concrete, when subjected to compression,
frequently give way by fracturing on one or more oblique planes
which cut the block into two wedges, or into pyramids and wedges.
The pyramids may overlap, and their bases are in the upper
and lower faces of the block. This mode of fracture, peculiar
to granular substances, of comparatively low shearing resistance,
can be discussed as follows:
If a short column, Fig. 2, of cross-section 5 is loaded cen-
trally with P, the unit compression on the right section will be
pi=P+S y and if the short column gives way under this load,
this value of pi is commonly considered the crushing strength
of the material. While it doubtless is the available crushing
strength of this specimen, it may by no means represent the maxi-
mum resistance to crushing under other conditions.
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i6
STRUCTURAL MECHANICS.
If pi=P +S is the unit thrust on the right section, it is seen,
from § 151, that, on a plane making an angle with the right
section, the normal unit stress ^n=^icos 2 0, and the tangential
unit stress q=pi sin 0cos0. If w= coefficient of frictional re-
sistance of the material to sliding, the resistance per square inch
1 ' S
Fig. 2
to sliding along this oblique plane will be tnp n =mpi cos 2 0, and
the portion of the unit shearing stress tending to produce fracture
along this plane will be q—mp n = pi (sin cos — m cos 2 0).
Fracture by shearing, if it occurs, will take place along that
plane for which the above expression is a maximum, or d(q—mp n )
+dQ=o. Differentiating relatively to 0,
^i(cos 2 —sin 2 + 2m sin cos 0) =0;
sin 2 — 2m sin cos + m 2 cos 2 0=cos 2 0+m 2 cos 2 0;
sin 0- m cos = cos 0\/(i +m 2 ) ;
sin „
H=tan 0=m+ v /(i +w 2 ).
COS v \ • /
If m were zero, max. would be 45°. Therefore the plane of
fracture always makes an angle greater than 45° with the right
section. As may be negative as well as positive, fracture tends
to form pyramids or cones.
Example. — A rectangular prism of cast iron, 2 in. high and square
section=i.os sq. in., sheared off under a load of 97,000 lb., or 92,380
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ACTION OF A PIECE UNDER DIRECT FORCE. 17
lb. compression per sq. in. of cross-section, at an angle whose tangent
was 1.5, or 56 19/.
i.5 = m+\/(i+ w2 ); 2.25— 3w+w 2 =i + w 2 ; m=o.42.
Sin 0=0.8321, cos ^==0.5546, sin cos 0=0.461, cos 2 0=0.308.
92,380(0.461— 0.42X0.308) =30,680 lb.
The coefficient of friction is 0.42, and the shear 30,680 lb. per sq. in.
The crushing strength of a short block would have exceeded consid-
erably the above 92,400 lb.
Since this deviation of the plane of fracture from 45 is due
to a resistance analogous to friction, it follows that, when a
column of granular material, and of moderate length, gives way
by shearing, the value pi will be only that compressive stress
which is compatible with the unit shearing strength, while its
real compressive strength in large blocks will be much higher.
The same phenomenon is exhibited by blocks of sandstone
and of concrete. Tests of cubes and flat pieces yield higher
results than do those of prisms of the same cross-section and
having a considerably greater height.
24. Ductile Substances under Compression. — Wrought iron,
and soft and medium steel, as well as other ductile substances,
tested in short blocks in compression, bulge or swell in trans-
verse dimensions, and do not fracture. Hence the ultimate
compressive strength is indefinite.
25. Fibrous Substances under Compression. — Wood and
fibrous substances which have but small lateral cohesion of the
fibres, when compressed in short pieces in the direction of the
same, separate into component fibres at some irregular section,
and the several fibres fail laterally and crush.
26. Vitreous Substances under Compression. — Vitreous sub-
stances, like glass and vitrified bricks, tend to split in the direc-
tion of the applied force.
27. Resistance of Large Blocks. — The resistance per square
inch of a cube to compression will depend upon the size of the
cube. As the unit stress and the resulting deformation are
associated, as noted in § 4, it follows that the unit compressive
stress will be greatest at the centre of the compressed surface
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1 8 STRUCTURAL MECHANICS.
and least at the free edges where lateral movement of the particles
is less restrained. Hence, the larger the cube, the greater the
mean or apparent strength per square inch. Large blocks of
stone, therefore, have a greater average sustaining power per
square inch than is indicated by small test specimens, other things
being equal.
The same inference can be drawn as to resistance of short
pieces to tension as compared with longer pieces of the same
cross-section.
A uniform compression over any cross-section of a large post
or masonry pier, when the load is centrally applied to but a small
portion of the top can be realized only approximately; the same
thing is probably true of the foundation below the pier. The
resisting capacity of the material, if earth, is thereby enhanced;
for the tendency to escape laterally at the edges of the foundation
is not so great as would be the case if the load were equally severe
over the whole base.
Beveling the edges of the compressed face of a block will
increase the apparent resistance of the material by taking the
load from the part least able to stand the pressure. The un-
loaded perimeter may then act like a hoop to the remainder.
Examples. — i. A round bar, i in. in diameter and 10 ft. long,
stretches 0.06 in., under a pull of 10,000 lb. What is the value of E?
What is the work done? 25,464,733; 300 in.-lb.
2. If the elastic limit of the bar is reached by a tension of 30,000
lb. per sq. in., what is the work done or the resilience of the bar?
1,666 in.-lb.
3. An iron rod, £=29,000,000, hangs in a shaft 1,500 ft. deep.
What will be the stretch? 1.55 in.
4. A certain rod, 22 ft. long, and having £=28,000,000, is to be
adjusted by a nut of 8 threads to the inch to an initial tension of 10,000
lb. per sq. in. If the connections were rigid, how much of a revolution
ought to be given to the nut after it fairly bears? 0.75.
5. Can a weight of 20,000 lb. be lifted by cooling a steel bar, 1 in.
sq., from 2i2 p to 62 F. ? Coefficient of expansion =0.001 2 for 180 ;
£=29,000,000.
6. A steel eye-bar, 80 in. long and 2 in. sq., fits on a pin at each
end with V in- play. What will be the tension in the bar, if the
temperature falls 75 F. and the pins do not yield?
7,250 lb. per sq. in.
7. A cross-grained stick of pine, 1 sq. in. in section, sheared off at
an angle of about 66 p with the right section under a compressive load
of 3,200 lb. If the coefficient of friction is 0.5, what is the unit shearing
stress of the section, the actual irregular area being 2.9 sq. in. ?
780 lb.
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CHAPTER n.
MATERIALS.
28. Growth of Trees. — Trees from which lumber is cut grow
by the formation of woody fibre between the trunk and the bark,
and each annual addition is more or less distinctly visible as a
ring. Each ring is made up of a light, porous part, the spring
wood, and a darker, dense part, the summer wood. Since the
latter is firm and heavy it determines to a large extent the weight
and strength of the timber. The sap circulates through the newer
wood, and in most trees the heart-wood, as it is called, can be
easily distinguished from the sap-wood. The former is con-
sidered more strong and durable, unless the tree has passed its
prime. The heart then deteriorates. Sap-wood, in timber ex-
posed to the weather, is the first to decay.
Branches increase in size by the addition of rings, as does
the trunk; hence a knot is formed at the junction of the branch
with the trunk. The knot begins where the original bud started,
and increases in diameter towards the exterior of the trunk, as
the branch grows. The grain of the annual growth, formed
around the junction of the branch with the trunk, is much dis-
torted. Hence timber that contains large knots is very much
weaker than straight-grained timber. Even small knots deter-
mine the point of fracture when timber is experimentally tested
for strength. When a branch happens to die, but the stub re-
mains, and annual rings are added to the trunk, a dead or loose
knot occurs in the sawed timber; such a knot is considered a
defect, as likely to let in moisture and start decay.
As forest trees grow close together, the branches die succes-
sively from below from lack of sunlight; such trees develop
19
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20 STRUCTURAL MECHANICS.
straight trunks of but little taper, free from any knots, except
insignificant ones immediately around the centre, and yield
straight-grained, clear lumber. A few trees, like hemlock, some-
times have their fibres running in a spiral, and hence yield cross-
grained timber. Trees that grow in open spaces have large side
limbs, and the lumber cut from them has large knots.
29. Shrinkage of Timber. — If a log is stripped of its bark
and allowed to dry or season, it will be found that the contrac-
tion or shrinkage in the direction of the radius is practically
nothing. There are numerous bundles or ribbons of hard tissue
running radially through the annual rings which appear to pre-
vent such shrinkage. Radial cracks, running in to a greater or
less distance, indicate that the several rings have yielded to the
tension set up by the tendency to shrink circumferentially. Sawed
timber of any size is likely to exhibit these season cracks. Such
cracks are blemishes and may weaken the timber when used
for columns or beams. By slow drying, and by boring a hole
through the axis to promote drying within, the tendency to form
season cracks may be diminished.
A board sawed radially from a log will not shrink in width,
and will resist wear in a floor. Such lumber is known as quarter-
sawed. A board taken off near the slab will shrink much and
will tend to warp or become concave on that side which faced
the exterior of the log. For that reason, and because the annual
rings have less adhesion than the individual fibres have, all boards
exposed to wear, as in floors, should be laid heart-side down.
30. Decay of Wood. — Timber exposed to the weather should
be so framed together, if possible, that water will not collect in
joints and mortises, and that air may have ready access to all
parts, to promote rapid drying after rain. The end of the grain
should not be exposed to the direct entrance of water, but should
be covered, or so sloped that water can run off, and the ends
should be stopped with paint. It is well to paint joints before
they are put together.
The decay of timber is due to the presence and action of
vegetable growths or fungi, the spores of which find lodgment
in the pores of the wood, but require air and moisture, with a
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MATERIALS. 21
suitable temperature, for their germination and spread. Hence
if timber is kept perfectly dry it will last indefinitely. If it is
entirely immersed in water, it will also endure, as air is excluded.
Moisture may be excluded from an exposed surface by the use
of paint. Unseasoned timber painted, or placed where there is
no circulation of air, will dry-rot rapidly in the interior of the
stick; but the exterior shell will be preserved, since it dries out
or seasons to a little depth very soon.
The worst location for timber is at or near the ground surface;
it is then continually damp and rot spreads fast.
31. Preservation of Wood. — The artificial treatment of timber
to guard against decay may be briefly described as the intro-
duction into the pores of some poison or antispetic to prevent
the germination of the spores; such treatment is efficacious as
long as the substance introduced remains in the wood. Creosote
is the best of preservatives and the only one effective against sea-
worms, but is expensive.
The timber is placed in a closed tank and steam is admitted
to soften the cells. After some time the steam is shut Qff, a
partial vacuum is formed, and the preservative fluid is run in and
pressure applied to force the liquid into the pores of the wood.
As steaming injures the fibres, treated timber is weaker than
untreated.
Burnettizing is the name given to treatment with zinc chloride,
a comparatively cheap process, applied to railway-ties and paving-
blocks. To prevent the zinc chloride from dissolving out in
wet situations, tannin has been added after the zinc, to form
with the vegetable albumen a sort of artificial leather, plugging
up the pores; hence the name, zinc-tannin process. For bridge-
timbers burnettizing makes the timber unduly brittle.
As the outside of treated timber contains most of the preserv-
ative, timber should be framed before being treated.
32. Strength of Timber. — The properties and strength of
different pieces of timber of the same species are very variable.
Seasoned lumber is much stronger than green, and of two pieces
of the same species and of the same dryness, the heavier is the
stronger, while, in general, heavy woods are stronger than light.
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STRUCTURAL MECHANICS.
Prudence would dictate that structures should be designed for
the strength of green or moderately seasoned timber of average
quality. As the common woods have a comparatively low re-
sistance to compression across the grain, particular attention
should be paid to providing sufficient bearing area where a
strut or post abuts on the side of another timber. An indenta-
tion of the wood destroys the fibre and increases the liability to
decay, if the timber is exposed to the weather, especially under
the continued working produced by moving loads.
Average breaking stresses of some American timbers as found
by the Forestry Div. of the U. S. Dept. of Agriculture follow.
The results are for well-seasoned lumber, 12 per cent, moisture:
Long-leaf pine. . .
Short -leaf pine. . .
White pine
Douglas spruce. .
White oak
Weight
Compression.
Bending.
per
Cu. Ft.
With
Grain.
Across
Grain.
Modulus
of
Rupture.
Modulus
of
Elasticity.
38
8,000
1,260
12,600
2,070,000
32
6,500
1,050
10,100
1,680,000
24
32
5»4QO
5,700
700
800
7,900
7,000
1,390,000
1,680,000
50
8,500
2,200
13,100
2,000,000
Shear
with
Grain.
835
770
400
500
i, 000
See also § 145.
Timber is graded or classified at the sawmills according
to the standard rules of different manufacturers' associations,
and specifications should call for a grade of lumber agreeing
with the classification of the mills of the region where the lumber
is produced.
33. Iron and Carbon. — Cast iron and steel differ from each
other in physical qualities on account of the different percentages
of carbon in combination with the iron. Ordinary cast or pig
iron contains from 3J to 4 per cent, of carbon, while structural
steel contains from one to two tenths of one per cent. Wrought
iron and steel are made by removing the metalloids from cast
iron.
34. Cast Iron. — Iron ore, which is an oxide of iron, is put in
a blast-furnace together with limestone and coke. Superheated
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MATERIALS. 23
air is blown in at the bottom of the furnace and the burning of
the coke produces a high temperature and removes the oxygen
from the ore. The earthy materials in the ore unite with the
limestone and form a slag which floats on the surface of the
molten metal and is drawn off separately. The iron is run off
into molds and forms pig iron.
When broken, the pig is seen to be crystalline and its color
may be white or gray, depending upon the condition of the carbon
in the iron. In the furnace the carbon is dissolved in the bath,
but when the iron solidifies the carbon may either remain in solu-
tion and produce white iron, or part of the carbon may precipitate
in the form of scales of graphite and produce gray iron. The
condition of the carbon depends partly on the rate of cooling,
but more on the other elements present. White iron is hard
and brittle; gray iron is tougher. If gray iron is run into a mold
lined with iron, it is chilled from the surface to a depth of one-
half to three-fourths of an inch; that is, the surface is turned
to white iron and made intensely hard, as in the treads of car-
wheels.
Besides the carbon, pig iron contains more or less silicon,
usually from one to two per cent. It tends to make the carbon
take the graphitic form. Sulphur makes the iron hard and
brittle; good foundry iron should not contain more than 0.15
of one per cent. Phosphorus makes molten iron fluid, and
irons high in phosphorus are used in making thin and intricate
castings, but such castings are very brittle. The amount of
silicon and sulphur in pig iron can be controlled by the furnace-
man, but the only way in which the amount of phosphorus can
be kept down is by using pure ores and fuel. The phosphorus
in pig iron to be used for making steel by the acid Bessemer
process is limited to one-tenth of one per cent.; iron fulfilling
that requirement is called Bessemer pig.
The tensile strength of cast iron varies from 15,000 to 35,000
lb. and the compressive strength from 60,000 to 200,000 lb. per
sq. in. The modulus of elasticity ranges from 10 to 30 million
pounds per square inch. For ordinary foundry iron the tensile
strength is usually 18,000 to 22,000 lb. and the modulus of elas-
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24 STRUCTURAL MECHANICS.
ticity 12 to 15 million. As cast iron is brittle and likely to contain
hidden defects, it is little used in structural work.
35. Wrought Iron. — Wrought iron is made by melting pig
iron, and cinder which contains oxide of iron, together in a rever-
beratory furnace. The carbon and silicon in the iron unite with
the oxygen of the slag, leaving metallic iron. As the carbon is
removed the melting-point rises, and since the temperature of the
furnace is not high enough to keep the iron fluid, it assumes a pasty
condition. The semi-fluid iron is collected into a lump by the
puddler and withdrawn from the furnace. It is then much like
a sponge; the particles of wrought iron have adhered to one
another, but each particle of iron is more or less coated with a
thin film of slag and oxide, as water is spread through the pores
of a partly dry sponge.
The lump of iron is put into a squeezer, and the fluid slag
and oxide drip out as water does from a squeezed sponge. But,
as it is impracticable to squeeze a sponge perfectly dry, so it is
impracticable to squeeze all the impurities out from among the
particles of metallic iron. In the subsequent processes of rolling
and re-rolling each globule of iron is elongated, but the slag
and oxide are still there; so that the rolled bar consists of a col-
lection of threads of iron, the adhesion of which to each other
is not so great as the strength of the threads.
If the surface of an iron bar is planed smooth and then etched
with acid, the metal is dissolved from the surface and the black
lines of impurities are left distinctly visible.
That wrought iron is fibrous is then an accident of the process
of manufacture, and does not add to its strength. If these im-
purities had not been in the iron when it was rolled out, it
would have been more homogeneous and stronger. The fibrous
fracture of a bar which is nicked on one side and broken by
bending is not especially indicative of toughness; for soft steel
is tough and ductile without being fibrous.
The tensile strength of wrought iron is about 50,000 lb. per
sq. in. and its modulus of elasticity about 28,000,000. Wrought
iron is still used to some extent when it is necessary to weld the
material, but soft steel has largely driven it from the market.
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MATERIALS. 25
36. Steel. — Structural steel is made from pig iron by burning
out the metalloids.
In the Bessemer process molten pig iron is run into a converter
and cold air is blown through the metal to burn out the carbon
and silicon. The combustion generates enough heat to keep
the metal fluid, although the melting-point rises. When the
metal is free from carbon and silicon, manganese is added to
remove the oxygen dissolved in the metal and to make the steel
tough when hot. If it is desired to add carbon to make a stronger
steel, spiegel-iron is used, which is a pig iron containing about
12 per cent, manganese and 4 or 5 per cent, carbon. After the
manganese is added the metal is cast into an ingot.
In America the acid Bessemer process is used exclusively;
that is, the converters are lined with a silicious material which
necessitates a silicious or acid slag, and consequently there is no
elimination of sulphur and phosphorus, which can enter a basic
slag only. This process is the cheapest way of making steel, but
the product is not so uniform as that of the open-hearth furnace.
Bessemer steel is used for rails and for buildings, but is not used
for first-class structural work.
An open-hearth furnace is a regenerative furnace having a
hearth for the metal exposed to the flame. Heat is generated
by burning gas which, together with the air-supply, has been
heated before it is admitted to the furnace. An intense heat is
thus produced which is sufficient to keep the metal fluid after
the carbon has been removed. The furnace is charged with
pig iron and scrap and, after the charge is melted, iron ore is
added. The carbon and silicon unite with the oxygen of the air
and of the ore. When the required composition is attained
the steel is drawn off, ferromanganese is thrown into the ladle
and the ingot is cast. After removal from the molds the ingots
are heated and rolled into plates or shapes. Steel castings are
made by pouring the metal from the furnace or the converter
into molds.
If the hearth is lined with sand, the slag formed during the
oxidation is silicious or acid and the oxide of phosphorus, which
acts as an acid, reunites with the iron. If, however, the hearth
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STRUCTURAL MECHANICS.
is lined with dolomite, limestone may be added to the charge
to form a basic slag which the phosphorus may enter. The
sulphur also may be reduced to some extent by the basic process,
but not by the acid.
The tensile strength of pure iron is probably about 40,000 lb.
per sq. in., but the presence of other elements, as carbon and
phosphorus in small quantities, increases its strength and makes
it more brittle. The element which increases the strength most
with the least sacrifice of toughness is carbon, and it is the element
which the manufacturer uses to give strength. In structural
steels it may range from 0.05 to 0.25 of one per cent. Phosphorus
and sulphur are kept as low as possible. Phosphorus makes
the steel brittle at ordinary temperatures, while sulphur makes
it brittle at high temperatures and likely to crack when rolled.
Manganese makes the steel tough while hot. It ranges from
0.30 to 0.60 of one per cent, in ordinary structural steels.
The strength of rolled steel depends somewhat upon the
thickness of the material, thin plates which have had more work
done upon them being stronger. The softer structural steels can
be welded readily, and the medium with care. They will not
temper. The modulus of elasticity of both soft and medium
steel is about 29,000,000 lb. per sq. in.
37. Classification of Steels. — The following classification and
requirements are taken from the standard specifications of the
American Society for Testing Materials.
Steel.
Tensile
Strength
in
1000 Lb.
Elonga-
tion in
8 In.
not Less
than
P
Acid
P
Basic
Mn
not more than
STRUCTURAL STEEL FOR BRIDGES AND SHIPS.
Rivet. . .
Soft
Medium
Rivet. . .
Medium
50 to 60
52 to 62
60 to 70
26%
25%
22%
0.08%
0.08%
0.08%
0.06%
0.06%
0.06%
0.06%
0.06%
0.06%
STRUCTURAL STEEL FOR BUILDINGS.
50 to 60
60 to 70
26%
O.IO%
O.IO%
O.IO%
O.IO%
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MATERIALS.
BOILER-PLATE AND RIVET-STEEL.
27
Steel.
Tensile
Strength
in
1000 Lb.
Elonga-
tion in
8 In.
not Less
than
P
Acid
P
Basic
S
Mn
not more than
Extra-soft
Fire-box
Boiler
45 to 55
52 to 62
55 to 65
28%
26%
25%
0.04%
0.04%
0.06%
0.04%
0.03%
0.04%
0.04%
0.04%
0.05%
0.30 to 0.50%
0.30 to 0.50%
. 30 to . 6o%
Steel for buildings may be made by the Bessemer process; the other two
classes must me made by the open-hearth process. Test specimens of rivet and
soft steel not more than three-fourths of an inch thick must bend cold 180 flat
without fracture; similar specimens of medium steel must bend cold 180 around
a diameter equal to the thickness of the piece without fracture. The yield-
point must not be less than one-half the ultimate strength.
This classification gives a good idea of the usual require-
ments for steel, although some engineers prefer a grade of
structural steel midway between soft and medium, that is, one
having a strength of 55 to 65 thousand pounds per square inch.
Such a grade is recommended by the American Railway
Engineering and Maintenance-of-Way Association.
38. Work of Elongation. — It is seen from the diagram Fig. 1
that the resistance of the metal per square inch increases as the
bar draws out, and diminishes in section under tension, as shown
by the dotted curve, although the total resistance grows less near
the dose of the test, as shown by the full line. As a small increase
in the amount of carbon diminishes the elongation and reduction
of area, it is possible that the carbon affects the apparent ultimate
strength in this manner (since such strength is computed on the
square inch of original section), and not by actually raising the
resisting power of the metal.
Since the measure of the work done in stretching a bar is the
product of one-half the force by the stretch, if the yield-point
has not been passed, and, for values beyond that point, is the
area below the curve in the diagram, limited by the ordinate
representing the maximum force, the comparative ability of
a material to resist live load, shock, and vibration is indicated
by this area. A mild steel of moderate strength may thus have
greater value than a higher carbon steel of much greater tensile
strength.
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28 STRUCTURAL MECHANICS.
39. Tool-steel. — Tool-steel as well as spring-steel of good
quality is made by melting wrought iron or steel of known compo-
sition in a crucible, and may contain from one-half to one per
cent, of carbon. When heated a bright red and quenched in water
such steel becomes very hard and brittle and entirely loses the
property of drawing out; but if it is subsequently heated to a
moderate temperature and then allowed to cool slowly, its
strength is increased and its brittleness reduced, while it still
retains more or less of its hardness. This process is called
tempering. Springs and tools are tempered before being used.
Some special tool-steels contain tungsten or chromium, which
give great hardness without tempering.
40. Malleable Iron: Case-hardened Iron. — There are two
other products which may well be mentioned, and which will be
seen to unite or fit in between the three already described. The
first is what is known as "malleable cast iron" or malleable iron.
Small articles, thin and of irregular shapes, which may be
more readily cast than forged or fashioned by a machine, and
which need not be very strong, are made of white cast iron, and
then imbedded in a substance rich in oxygen, as, for instance,
powdered red hematite iron ore, sealed up in an iron box, and
heated to a high temperature for some time. The oxygen ab-
stracts the carbon from the metal to a slight depth, converting
the exterior into soft iron, while the carbon in the interior takes
on a graphitic form with an increase of strength and diminution
of brittleness.
The second product is case-hardened iron. An article fash-
ioned of wrought iron or soft steel is buried in powdered charcoal
and heated. The exterior absorbs carbon and is converted into
high steel, which will better resist wear and violence than will
soft iron. The Harvey process for hardening the exterior of
steel armor-plates is of a similar nature.
41. Effect of Shearing and Punching. — As was shown in
§ 15, when a bar of steel is stressed beyond the elastic limit and
has received a permanent set, a higher elastic limit is established,
but the percentage of elongation is much reduced, as shown by
the curve E F N of Fig. 1. The steel therefore has been hardened
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MATERIALS. 29
in the sense that its ductility has been lessened. Examples of
this hardening are seen in plates which are rolled cold and in
drawn wire. Similarly when a rivet-hole is punched in a plate,
the metal immediately surrounding the hole is distorted and
hardened, thus reducing the ductility of the plate around the
hole. If the strip containing a punched hole is loaded beyond
the elastic limit of the plate, the metal surrounding the hole,
being unable to stretch as much as the rest, is unduly stressed
and the ultimate strength is less than it would have been had
the stress been uniformly distributed. Experiments show that
plates with punched holes are weaker than those with drilled holes.
The same hardening effect is produced by shearing, or cutting,
a plate. When a bar of punched or sheared steel is bent, cracks
form at the hard edges and spread across the plate; but if the
holes are reamed out or the sheared edges are planed off to a
small depth, the hardened metal is removed and the bar will
bend without cracking. Medium steel, especially if thick, is
injured much more than soft steel by punching and shearing.
Specifications for structural work frequently require rivet-holes to
be reamed to a diameter three-sixteenths of an inch larger than the
punch, and one-quarter of an inch to be planed from the edges of
sheared plates. In good boiler- work the rivet-holes are drilled.
The ductility of steel which has been hardened by cold work-
ing can be restored by annealing, that is, by heating to a red
heat and then cooling slowly.
42. Building Stone. — The principal building stones may
be grouped as granites, limestones, and sandstones. Granite
consists of crystals of quartz, felspar and mica or hornblende.
It is very strong and durable, but its hardness makes it difficult
to work. Owing to its composite structure it does not resist
fire well. Limestone is a stratified rock of which carbonate of
lime is the chief ingredient. When limestone consists of nearly
pure carbonate and is of good color and texture, it is called
marble. Sandstone consists of grains of sand cemented together
by silica, carbonate of lime, iron oxide, or clay. If the cementing
material is silica, the stone is very hard to work. Sandstone is
one of the most valuable of building materials-
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3° STRUCTURAL MECHANICS.
Sound, hard stones like granite, compact limestone, and the
better grades of sandstone are sufficiently strong to cany any
loads brought upon them in ordinary buildings; hence the question
of durability rather than strength is the governing consideration
in selecting a good building stone. The only sure test of the
ability of a building stone to resist climatic changes, to stand the
weather, is the lapse of time. Artificial freezing and thawing
of a small specimen, frequently repeated, will give indication as
to durability.
Stratified stones should be laid on their natural beds, that is, so
that the pressure shall come practically perpendicular to the layers.
They are much stronger in such a position, and the moisture
which porous stones absorb from the rain can readily dry out.
If the stones are set on edge, the moisture is retained and, in the
winter season, tends to dislodge fragments by the expansive force
exerted when it freezes. Some sandstone facings rapidly deterio-
rate from this cause. Crystals of iron pyrites occur in some sand-
stones and unfit them for use in the face of walls. The
discoloration resulting from their oxidation, and the local break-
ing of the stone from the swelling are objectionable.
The modulus of elasticity differs greatly for different stones.
Limestones and granites are nearly perfectly elastic for all work-
ing loads, but sandstones take a permanent set for the smallest
loads. Tests of American building stones in compression made at
the Watertown Arsenal give values of 5 to 10 million for granites
and marbles and 1 to 3 million for sandstones. The weight of
granite ranges from 160 to 180, of limestone and marble from
150 to 170, and of sandstone from 130 to 150 pounds per
cubic foot.
43. Masonry. — Most masonry consists of regularly coursed
stones on the face, with a backing of irregular-shaped stones
behind. Stones cut to regular form and laid in courses make
ashlar masonry, if the stones are large and the courses continuous.
When the stones are smaller, and the courses not entirely con-
tinuous, or sometimes quite irregular, although the faces are
still rectangular, the descriptive name is somewhat uncertain, *
as block-in-course, random range, etc., down to coursed rubble,
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MATERIALS. 31
where the end joints of the stones are not perpendicular to the
beds. Rubble masonry denotes that class where the stones are
of irregular shape, and fitted together without cutting. If the
face of the stone is left as it comes from the quarry, the work is
called quarry -faced or rock-faced. The kind of masonry depends
upon the beds and joints. Walls of stone buildings have only
a more or less thin facing of stone, the body of the wall being
of brick. The stone facing should be well anchored to the brick-
work by iron straps.
44. Bricks. — Bricks are made from clay which may be
roughly stated to be silicate of alumina (Al 2 C>3,2Si02,2H20).
The clay is freed from pebbles, mixed with water in a pug-mill
and molded. The green bricks are dried in the air and then
burned in kilns. Pressed bricks are pressed after drying. They
have a smooth exterior, are denser and are more expensive than
common bricks. Pairing-bricks are made from hard, laminated,
rock-like clays called shales, which are not plastic unless pulver-
ized and mixed with water. Paving- bricks are burned to incipient
vitrification, which makes them extremely hard. Lime and iron
in clay act as fluxes and make the clay fusible; fire- bricks are
therefore made from clay free from fluxes. If limestone pebbles
occur in a brick-clay, they must be removed or they will form
lumps of lime after burning, and when wet will slake, swell, and
break the bricks.
The red color of common bricks is an accidental character-
istic, due to iron in the clay. Such bricks are redder the harder
they are burned, finally, in some cases, turning blue. The cream-
colored bricks with no iron may be just as strong and are com-
mon in some sections. Soft, underburned bricks are very porous,
absorb much water, and cannot be used on the outside of a
wall, especially near the ground line, for they soon disintegrate
from freezing. Hard-burned bricks are very strong and satis-
factory in any place; they can safely carry six or eight tons to
the square foot. Bricks differ much in size in different parts
of the country. A good brick should be straight and sharp-
edged, reasonably homogeneous when broken, dense and heavy.
Two bricks struck together should give a ringing sound.
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32 STRUCTURAL MECHANICS.
Sand bricks are made by mixing thoroughly sand with
5 or 10 per cent, of slaked lime and sufficient water to allow
molding. The bricks are formed under very great pressure
and are then run into a large boiler and exposed to the action of
steam under pressure for several hours. Some chemical reaction
takes place between the silica and the lime under the conditions
of heat and moisture, which firmly cements the particles of sand.
Well-made sand bricks have a crushing strength of 2,500 to 5,000
lb. per sq. in. They are denser than common bricks and are
very regular in shape and size.
45. Lime. — Lime for use in ordinary masonry and brick-
work is made by burning limestone, or calcium carbonate, CaC0 3|
and thus driving off by a high heat the carbon dioxide and such
water as the stone contains. There remains the quicklime of com-
merce, CaO, in lumps and powder. This quicklime has a great
affinity for water and rapidly takes it up when offered, swelling
greatly and falling apart, or slaking, into a fine, dry, white powder,
Ca(OH)2, with an evolution of much heat, due to the combina-
tion of the lime with the water. The use of more water produces
a paste, and the addition of sand, which should be silicious,
sharp in grain and clean, makes lime mortar. The sand is
used partly for economy, partly to diminish the tendency to crack
when the mortar dries and hardens, and partly to increase the
crushing strength. The proportion is usually 2 or 2 \ parts by
measure of sand to one of slaked lime in paste, or 5 to 6 parts
of sand to one of unslaked lime. As lime tends to air-slake, it
should be used when recently burned.
Some limes slake rapidly and completely; other limes have
lumps which slake slowly and should be allowed time to com-
bine with the water. It is generally considered that lime mortar
improves by standing, and that mortar intended for plastering
should be made several days before it is used. Small unslaked
fragments in the plaster will swell later and crack the finished
surface. The lime paste is sometimes strained to remove such
lumps.
Lime mortar hardens by the drying out of part of the water
which it contains, and by the slow absorption of carbon dioxide
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MATERIALS. 3$
from the air. It thus passes back by degrees to a crystallized cal-
cium carbonate surrounding the particles of sand : Ca(OH) 2 +CO2
= CaC0 3 +H20. Dampness of the mortar is favorable to the
attainment of this result, and the mortar in a brick wall which
has been kept damp for some time will harden better than where
the wall is dry. Dry, porous bricks absorb rapidly, and almost
completely, the water from the mortar, and reduce it to a powder
or friable mass which will not harden satisfactorily. Hence
bricks should be well, wetted before they are laid.
Lime mortar in the interior of a very thick wall may not
harden for a long time, if at all, and hence should not be used
in such a place. Slaked lime placed under water will not harden,
as may be proved by experiment. In both cases such inaction
is due to the exclusion of the carbon dioxide. Lime mortar
should never be used in wet foundations.
Plaster for interior walls is lime mortar. Hair is added
to the mortar for the first coat, so that the portion which is forced
through the spaces between the laths and is clinched at the back
may have sufficient tenacity to hold the plaster on the walls
and ceiling.
46. Natural Cement. — Natural cement is made by burning
almost to vitrification a rock which contains lime, silica, and
alumina, that is, one which may be considered a mixture of a
limestone and a clay rock. The carbon dioxide, moisture, and
water of crystallization are expelled by burning. The hard
fragments must then be ground to powder, the finer the better.
If the rock contains the several ingredients in proper proportions,
the addition of water to the powder makes a plastic mass which
hardens or sets by crystallization. This setting may begin in a
few minutes or half an hour. The hardening, the tensile and
compressive strengths increase rapidly at first, and at a decreasing
rate for months.
As access of air is not required for the setting of cement,
the reaction taking place when water is added to the dry powder,
cement mortar is used invariably under water and in wet places.
It makes stronger work than lime mortar, and is generally used
by engineers for stone masonry. Its greater cost than that of
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34 STRUCTURAL MECHANICS.
lime is due to the necessity of grinding the hard clinker; while
lime falls to powder when wet. The proportion of sand is i, 2,
or 3 to one of cement, according to the strength desired, 2 to 1
being a common ratio for good work. The sand and cement are
mixed dry and then wetted, in small quantities, to be used at once.
The addition of brick-dust from well-burned bricks to lime
mortar will make the latter act somewhat like cement, or become
hydraulic, as it is called. Volcanic earth has been used in the
same way.
47. Portland Cement. — If the statement made as to the com-
position of cement is correct, it should be possible to make a
mixture of chalk, lime or marl, and clay in proper proportions
for cement, and the product ought to be more uniform in com-
position and characteristics than that from the natural rock.
Such is the case, and in practice about three parts of carbonate
of lime are intimately mixed with one part of clay and burned
in kilns. During the burning the combined water and carbon
dioxide are driven off and various compounds are formed of which
tricalcium silicate (3CaO,Si02) is the most important, as it is
the principal active element and constitutes the greater part of
hydraulic cements. The resulting clinker is ground to an impal-
pable powder which forms the Portland cement of commerce.
Fine grinding is essential, as it has been shown that the coarser
particles of the cement are nearly inert.
Upon mixing the cement with water, the soluble salts dis-
solve and crystallize, that is, the cement sets. The water does
not dry out during hardening as in lime mortar, but combines
with some of the salts as water of crystallization. This crystal-
lization takes place more slowly in the Portland than in the
natural cements, but after the Portland cement has set it is much
harder and stronger than natural cement. The slower-setting
cement mortars are likely to show a greater strength some months
or years after use than do the quick-setting ones, which attain
considerable strength very soon, but afterwards gain but little.
48. Cement Specifications. — The following specifications of
cement are reasonable:
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MATERIALS. 35
NATURAL CEMENT.
Specific gravity: not less than 2.8.
Fineness: 90 per cent, to pass a sieve of 10,000 meshes per
square inch.
Setting: initial set in not less than fifteen minutes; final set
in not more than four hours.
Soundness: thin pats of neat cement kept in air or in water
shall remain sound and show no cracks.
Tensile strength: briquettes one inch square in cross-section
shall develop after setting one day in air and the remaining time
in water:
Neat 7 days 100 lb.
" 28 " 200lb.
1 cement, 1 sand.. 7 " 60 lb.
.28 " 1501b.
a a
PORTLAND CEMENT.
Specific gravity: not less than 3.10.
Fineness: 92 per cent, to pass a sieve of 10,000 meshes per
square inch.
Setting: initial set in not less than thirty minutes; final set in
not more than ten hours. (If a quick-setting cement is desired
for special work, the time of setting may be shortened and the
requirements for tensile strength reduced.)
Soundness: a thin pat of neat cement kept in air 28 days
shall not crack; another pat allowed to set and then boiled for
five hours shall remain sound.
Tensile strength: briquettes, as for natural cement:
Neat 7 days 450 lb.
" 28 " 5501b.
1 cement, 3 sand. . 7 ' ' 150 lb.
11 ..28 " 200lb.
49. Concrete. — Concrete is a mixture of cement mortar (cement
and sand) with gravel and broken stone, the materials being so
proportioned and thoroughly mixed that the gravel fills the
spaces among the broken stone; the sand fills the spaces in the
gravel; and the cement is rather more than sufficient to fill the
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36 STRUCTURAL MECHANICS.
interstices of the sand, coating all, and cementing the mass into
a solid which possesses in time as much strength as many rocks.
It is used in foundations, floors, walls, and for complete structures.
The broken stone is usually required to be small enough to pass
through a 2-in. or 2^-in. ring. The stone is sometimes omitted.
To ascertain the proportions for mixing, fill a box or barrel
with broken stone shaken down, and count the buckets of water
required to fill the spaces; then empty the barrel, put in the
above number of buckets of gravel, and count the buckets of
water needed to fill the interstices of the gravel; repeat the
operation with that number of buckets of sand, and use an amount
of cement a little more than sufficient to fill the spaces in the sand.
If the gravel is sandy, screen it before using, in order to keep
the proportions true. A very common rule for mixing is one
part cement, three parts sand, and five parts broken stone or
pebbles, all by measure.
The ingredients are mixed dry, then water is added and the
mass is mixed again, after which it is deposited in forms in layers
6 or 8 inches thick. Experience has shown that a mixture wet
enough to flow makes a denser concrete than a dry mixture,
especially if the mass cannot be thoroughly tamped.
50. Paint. — When a film of linseed-oil, which is pressed from
flaxseed, is spread on a surface it slowly becomes solid, tough,
and leathery by the absorption of oxygen from the air. In order
that the film may solidify more rapidly the raw oil may be pre-
pared by heating and adding driers, oxides of lead and man-
ganese, which aid the oxidation; oil treated in this way is called
boiled oil. Driers should be used sparingly, as they lessen the
durability of paint. An oil-film is somewhat porous and rather
soft, hence its protective and wearing qualities can be improved
by the addition of some finely ground pigment to fill the pores
and make the film harder and thicker. Most pigments are inert.
Paint, then, consists of linseed-oil, a pigment, and a drier. Varnish
is sometimes added to make the paint glossy and harder, or tur-
pentine may be used to thin it
Varnishes are made by melting resin (resins are vegetable
gums, either fossil or recent), combining it with Unseed -oil, and
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MATERIALS. 37
thinning with turpentine. They harden by the evaporation of
the turpentine and the oxidation of the oil and resin. The
addition of a pigment to varnish makes enamel- or varnish-
paint.
For painting on wood white lead, the carbonate, and white
zinc, an oxide, are pigments extensively used. Iron oxide is
largely used on both wood and steel. Red lead, an oxide, and
graphite are pigments used on steel. Red lead acts differently
from other pigments in that it unites with the oil, and the mixture
hardens even if the air is excluded, so that red-lead paint must
be mixed as used. Lampblack is often mixed with other pig-
ments to advantage, or it is sometimes used alone.
As paint is used to form a protective coating, it should not
be brushed out too thin, but as heavy a coat as will dry uniformly
should be applied. Wood should be given a priming coat of
raw linseed-oil, so that the wood shall not absorb the oil from the
first coat of paint and leave the pigment without binder. In
applying paint to steel-work it is essential for good work that
the paint be spread on the clean, bright metal. Rust and mill-
scale must be removed before painting if the coat is expected
to last. As mill-scale can be removed only by the sand-blast
or by pickling in acid, steel is seldom thoroughly cleaned in
practice. If paint is applied to rusty iron, the rusting will go on
progressively under the paint. Painting should never be done in
wet or frosty weather.
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CHAPTER m.
BEAMS.
Si. Beams: Reactions. — A beam may be defined to be a
piece of a structure, or the structure itself as a whole, subjected
to transverse forces and bent by them. If the given forces do
not act at right angles to the axis or centre line of the piece, their
components in the direction of the axis cause tension or com-
pression, to be found separately and provided for; the normal or
transverse components alone produce the beam action or bending.
As all trusses are skeleton beams, the same general principles
apply to their analysis, and a careful study of beams will throw
much light on truss action.
Certain forces are usually given in amount and location on
a beam or assumed. Such are the loads concentrated at points
or distributed over given distances, and due to the action of
gravity; the pressure arising from wind, water, or earth; or the
action of other abutting pieces.
It is necessary, in the first place, to satisfy the requirements
of equilibrium, that the sum of the transverse forces shall equal
zero and that the sum of their moments about any point shall
also equal zero. This result is accomplished by finding the
magnitudes and direction of the forces required at certain given
points, called the points of support, to produce equilibrium.
The supporting forces or reactions, exerted by the points of support
against the beam, are two or more, except in the rare case where
the beam is exactly balanced on one point of support. For cases
where the reactions number more than two, see § 109.
52. Beam Supported at Two Points. Reactions. — The
simplest and most generally applicable method for finding one
38
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BEAMS. 39
of the two unknown reactions is to find the sum of the moments
of the given forces about one of the points of support, and to
equate this sum with the moment of the other reaction about the
same point of support. Hence, divide the sum of the moments
of the given external forces about one of the points of support
by the distance between the two points of support, usually called
the span, to find the reaction at the other point of support. The
direction of this reaction is determined by the sign of its moment,
as required for equilibrium. The amount of the other reaction
is usually obtained by subtracting the one first found from the
total given load.
W 'f )0 ,W-750 IK-150 -600-R,
-ft 3— a. 4 V-t — d 1 zn r * — "
^*"° r te .. «»-* ^i * Fig.5 *«
CB
Thus, in the three cases sketched, P\ = W-jr^\ P 2 =W—P\.
Examples.— Fig. 3. If W= 500 lb., A B = 30 ft., and B C = 18 ft. ;
D 500-18
-ri = - =300 lb., i > 2= 500—300= 200 lb.
Fig. 4. K ^=750 lb., A B=2o ft., and A C=5 ft., P^J^IS-
20
937J lb., and P 2 =75o-937i=-i*7l lb.
Fig. 5. If W=i S o lb., A C=2o ft, and A B = 5 ft., P^&IH*.
750 lb., and P 2 = 150— 750= — 600 lb. Note the magnitude of P\ and
P2 as compared with W when the distance between P\ and P 2 is small.
Such is often the case when the beam is built into a wall.
Where the load is distributed at a known rate over a certain
length of the beam, the resultant load and the distance from its
point of application to the point of support may be conveniently
used.
♦•#-- i~»-+— «*• — if— «i» y a .1? 00 «°
RRTT^T
>-~ it—— 4 • c A 7 o" e r— B .
?> • mt* ^— * ~ : ^t
Pig. 6. F, «- 7
Example.— Tig. 6. If AB=40 ft, AD=8 ft, DE=i6 ft, and
the load on D E is 200 lb. per ft, ^=3,200 lb., and CB=24 ft.
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40
STRUCTURAL MECHANICS.
1 200 * 2 A.
Therefore Pi = — =1,920 lb., and P 2 = 3, 200— 1,920=1,280 lb.
40
If several weights are given in position and magnitude, the
same process for finding the reactions, or forces exerted by the
points of support against the beam, is applicable.
Examples. — In Fig. 7, Pi = (ioo-i8-f 200- 16-f 150- 13 +300-11-1-
50-8+ 80.0)4-16 = 665! lb. P 2 =88o-66st=2i4f lb. The work can
3Jj +
Fig. 10
9
1000
:j
ii-200
be checked by taking moments about A to find P 2 , the moment 100*2
then being negative.
If the depth of water against a bulkhead, Fig. 8, is 9 ft., and the
distance between A and B, the points of support, is 6 ft., A being at
the bottom, the unit water pressure at A will be 9X62.5 = 562.5 lb.
which may be represented by A D, and at other points will vary with
the depth below the surface, or as the ordinates from E A to the inclined
line E D. Hence the total pressure on E A, for a strip 1 ft. in hori-
zontal width, will be 562.5X94-2=2,531^ lb., and the resultant pres-
sure will act at C, distant | A E, or 3 ft. from A. P 2 = 2,531^X3 4-6=
1,265.6 lb., and Pi = 2,531.2— 1,265.6=1,265.6 lb., a result that might
have been anticipated, from the fact that the resultant pressure here
passes midway between A and B.
Let 1,000 lb. be the weight of pulley and shaft attached by a hanger
to the points D and E, Fig. 9. Let the beam A B= 10 ft., A D=4 ft.,
D E=4 ft., E B=2 ft.; and let C be 2 ft. away from the beam. As
the beam is horizontal, Pi = 1,000X44-10=400 lb.; P 2 =i,ooo— 400=
600 lb., and both act upwards. The 1,000 lb. at C causes two vertical
downward forces on the beam, each 500 lb., at D and E. There is also
compression of 500 lb. in D E.
When the beam is vertical, Fig. 10, by moments, as before, about
B, Pi= i,ooo-2-r- 10= 200 lb. at A acting to the left, being tension or
a negative reaction. By moments about A, P 2 = 1,000-2 4- 10=200
at B, acting to the right. Or Pi+P 2 =o; . • . P lS = -P 2 . By similar
moments, the 1,000 lb. at C causes two equal and opposite horizontal
forces on the beam at D and E, of 500 lb. each, that at D being ten-
sion on the connection, or acting towards the right, and that at E
acting in the opposite direction. These two forces make a couple
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MATERIALS. 41
balanced by the couple P\P2- The weight 1,000 lb. multiplied by
its arm 2 ft. is balanced by the opposing horizontal forces at D and E,
4 ft. apart. There remains a vertical force of 1,000 lb. in A B, which
may all be resisted by the point B, when the compression in DE=
500 lb. and in E B= 1,000 lb.; or all by the point A, when the tension
in D E=5oo lb. and in D A= 1,000 lb.; or part may be resisted at A,
and the rest at B, the distribution being uncertain. This longitudinal
force may be disregarded in discussing the beam, as may the tension
or compression in the hanger arms themselves.
53. Bending Moments. — If an imaginary plane of section
is passed through any point in a beam, the sum of the moments
of all the external forces on one side of that section, taken about
a point in the section, must be exactly equal and opposite to
the sum of the moments of all the external forces on the other
side of that section, taken about the same point. If not, the beam
would revolve in the plane of the forces. The moment on the
left side of the section tends to make that portion of the beam
rotate in one direction about the point of section, and the equal
moment on the right side of the section tends to make the right
segment rotate in the opposite direction. These two moments
cause resistances in the interior of the beam at the section (which
stresses will be discussed under resisting moment), with the result
that the beam is bent to a slight degree. Either resultant moment
on one side of a plane of section, about the section, is called the
bending moment at that point, usually denoted by M , and is con-
sidered positive when it makes the beam concave on the upper
side. Ordinary beams, supported at the ends and carrying
loads, have positive bending moments.
If upward reactions are positive, weights must be taken as
negative and their sign regarded in writing moments.
Examples.— Section at D, Fig. 3, 10 ft. from B. On the left of D,
and about D, Pi (=300) -20— 500-8=2,000 ft. -lb., positive bending
moment at D. Or, about D, on the right side of the section;
P 2 (= 200) • 10= 2,000 ft.-lb., positive bending moment at D. Usually
compute the simpler one.
Section at A, Fig. 4, WC A= -750- 5= -3,750 ft.-lb. negative
bending moment at A, tending to make the beam convex on the upper
side. At D, 10 ft. from B, M = -P 2 - 10= -- 187$- 10= - 1,875 ft.-lb.,
negative because P2 is negative. At A, taking moments on the right
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42 STRUCTURAL MECHANICS.
of and about A, Af= — i87£- 20=— 3,750 ft.-lb., as first obtained.
This beam has negative bending moments at all points.
In Fig. 5, M at D is— 150-10= — 1,500 ft.-lb. It is evident that
the bending moments at all points between C and A can be found
found without knowing the reactions. If this beam is built into a
wall, the points of application of P\ and P 2 are uncertain, as the pres-
sures at A and B are distributed over more or less of the distance that
the beam is embedded. The maximum M is at A, and is — 150- 20=
— 3,000 ft.-lb. It is evident that the longer A B is, the smaller the reac-
tions are, and hence the greater the security.
In Fig. 6, the bending moment at C will be Pi -AC— weight on
D C-iD C= 1,920- 16-200-8-4=24,320 ft.-lb. At E, ^ = 1,280-16=
.20,480 ft.-lb.
In Fig. 7, the bending moments at the several points of application
of the weights, taking moments of all the external forces on the left of
each section about the section, will be —
At C, M= — ioo-o=o.
At A, M= — ioo- 2= — 200 ft.-lb.
At D, M= - 100 • 5 + (665!- 200) -3 =896} ft.-lb.
At E, M=- 100-7 +465$ 5- 150- 2 = 1,328$ ft.-lb.
At F, M=- ioo- 10+4651-8- 150- 5-300-3 = 1,075 ft.-lb.
And, at B, M will be zero. M max. occurs at E.
Do not assume that the maximum beading moment will
be found at the point of application of the resultant of the .load.
The method for finding the point or points of maximum bending
moment will be shown later.
The moments on the right portion of the beam may be more
easily found by taking moments on the right side of any section.
Thus at F, Jkf = (P 2 -8o)-8 = (2i4f-8o)-8 = i,o75 ft.-lb. Find
the bending moment at the middle of E F. i,20iA ft.-lb.
In Fig. 8, the bending moment at section C of the piece A E may
be found by considering the portion above C. As the unit pressure
at C is 6X62$ lb. = 375 lb. per sq. ft., M at C= P 2 (= 1,265.6) -3
— (375X64-2)-6-^3 = 1,546.8 ft.-lb. At the section B, Jf=
-(3X62*X3-2)Xi = -28iift.4b.
In Fig. 9, as P\ = 400 lb., P 2 =6oo lb., vertical forces at D and E
are each 500 lb.; M at D= 1,600 ft.-lb.; M at E= 1,200 ft.-lb.
In Fig. 10, as Pi = — 200 lb. = — P 2 , and the horizontal forces at D
and E are ± 500 lb.; M at D=-8oo ft.-lb.; M at E=+4cx> ft.-lb.
The beam will be concave on the left side at D and convex at E. The
curvature must change between D and E, where M=o. Let this
point be distant x from B. Then 200-*— 5oo(#— 2)=o; .\ x=>
3i».
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BEAMS 43-
The curved piece A B, Fig. 11; with equal and opposite forces
applied in the line connecting its ends, will experience a bending
moment at any point D, equal to P-CD, this ordinate being
perpendicular to the chord.
54. Shearing Forces. — In Fig. 3, of the 500 lb. at C, 300 lb.
goes to A and 200 lb. to B. Any vertical section between A and C
must therefore have 300 lb. acting vertically in it. On the le**t
of such a section there will be 300 lb. from Pi acting upwards,
and on the right of the same section there will be 300 lb., coming
from W, acting downwards. These two forces, acting in opposite
directions on the two sides of the imaginary section, tend to cut
the beam off, as would a pair of shears, and either of these two
opposite forces is called the shearing force at the section, or simply
the shear. When acting upwards on the left side of the section
(and downwards on the right side), it is called positive shear.
When the reverse is the case the shear will be negative.
Examples. — In Fig. 7, where a number of forces are applied to a
beam, there must be found at any section between C and A a shear
of —100 lb.; between A and D the shear will be — 100+6651—200=
+365$ lb.; between D and E the shear will decrease to 3651—150=
21 5f lb.; on passing E the shear will change sign, being 2151—300=
— 84I lb.; between F and B it will be — 841—50= — 134$ lb.; and
on passing B, it becomes zero, a check on the accuracy of the several
calculations.
In Fig. 8, the shear just above the support 6=3X62^X3^2 =
281} lb. ; just below the point B the shear is 281}— 1,265.6= —984.4 lb. ;
and just above A it is 1,265.6 lb. The signs used imply that the left
side of A E corresponds to the upper side of an ordinary beam. As
the shear is positive above A and negative below B, it changes sign
at some intermediate point. Find that point.
In Fig. 9, the shear anywhere between A and D is +400 lb.; at
all points between D and E it is 400— 500= — 100 lb.; and between
E and B is —600 lb. The shear changes sign at D.
In Fig. 10, the shear on any horizontal plane of section between
B and E is —200 lb.; betwen E and D is —200+ 500= +300 lb.;
and between D and A is +300— 500= — 200 lb. The shear changes
sign at both E and D.
55. Summary. — To repeat: — The shearing force at any
normal section of a beam may therefore be defined to be the
algebraic sum 0} all the transverse forces on one side of the section.
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44 STRUCTURAL MECHANICS.
When this sum or resulting force acts upward on the left of the
section, call it positive; when downward, negative.
The bending moment at any right or normal section of a beam
may be stated to be the algebraic sum of the moments oj all the
transverse forces on one side oj the section, taken about the centre
oj gravity oj the section as axis. When this sum or resulting
moment is right-handed or clockwise on the left of and about
the section, call it positive. A positive moment tends to make
the beam concave on what is usually the upper side.
By a proposition in mechanics, any force which acts at a
given distance from a given point is equivalent to the same force
at the point and a moment made up of the force and the perpen-
dicular from the point to the line of action of the force. Then
in Fig. 7, if a section plane is passed anywhere, as between D
and E, the resultant force on the left, which is the algebraic sum
of the given forces on the left of the section, is the shear at the
section; and this resultant, multiplied by its arm or distance
from the point in D E, giving a moment which is the algebraic
sum of the moments of the several forces on the left of and about
the point, is the bending moment at the section.
It is also evident that the resulting action at any section is
the sum of the several component actions; and hence that dif-
ferent loads may be discussed separately and their effects at
any point added algebraically, if they can occur simultaneously.
Thus the shears and bending moments arising from the weight
of a beam itself may be determined, and to them may be added
the shears and bending moments at the same points from other
weights imposed on the beam.
The numerous examples already given show that formulas
are not needed for solving problems in beams, and the student
will do well to accustom himself to using the data directly.
Formulas, however, will now be derived, which will sometimes
be convenient for use, and from which may be deduced certain
serviceable relationships.
56. Bending Moment a Maximum where the Shear Changes
Sign. — If a beam weighing w per unit of length is supported
at each end and carries a system of loads any one of which is
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BEAMS. 45
distant a from the left support, the shear at a section distant x
from the left support is
Fx^P^IlW-wx,
and the bending moment at the same section is
M x =P x x-2o W(x-a) -\w*?.
If the beam is a cantilever, that is, a beam fixed in position
at the right end and unsupported at the left, the same equation
will apply when P x becomes zero. It is seen by comparing the
equations above that F is always the first derivative of M, or
dM x
Hence, according to the rule for determining maxima and minima,
the bending moment is always a maximum (or minimum) at the
place where the shear is zero or changes in sign. This criterion
is easily applied to locate the points of M maximum. Pass
along the beam from the left (or right) until as much load is on
the left (or right) of the section as will neutralize P\ (or P 2 ) and
the point of M max. is found. Its value can then be computed.
If the weight at a certain point is more than enough to reduce
F to zero, F changes sign in passing that point, and hence M
max. occurs there.
For a beam fixed at one end only, F changes sign in passing
Pi, and hence M max. is found at the wall.
Examples. — M max. occurs in Fig. 3, at C; in Fig. 4, at A; in
Fig. 6, at 17.6 ft. from A; in Fig. 7, at A, and again at E; in Fig. 8,
at B, and again at a distance x from E such that 62 J* • %x= 1,265$; • ' • x
=V / 40-5 =s 6.36 ft. ; in Fig. 9, at D; and in Fig. 10, at D and again at E.
The bending moments which may not have been found at some of
these points can now be computed.
The reader who is familiar with graphics can draw the equi-
librium or bending-moment polygons or curves, and the shear
diagrams, and notice the same relation in them.
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46 STRUCTURAL MECHANICS.
The unit had may also be considered as the derivative of
the shear; F therefore has maximum (or minimum) values where
the external forces change in sign.
The origin of coordinates may be arbitrarily taken at any
point in the length of the beam and general expressions may
be written. If —w is the unit load and is constant,
F 9 = — f*wdx=Fo—wx,
M x = f*F:dx =M Q +Fox- \wx?,
in which Fq and Mo are the constants of integration, the values of
F and M at the origin. Thus in the beam above, the bending
moment at a second section distant c from the first is
M x ^=M x +F x c~r x +e W(x-a+c)-$we*.
The same expression is easily derived by substituting x+c
for x in the original equation.
Example— In Fig. 7, M at D= + 8q6| ft.-lb. F between D and E=
+ 215$ lb. M at E= +8961+2151X2=1,328$ ft.-lb. as in §53.
Mmidway between E and F=896$+2i5fX3£— 300X1$= 1, 201ft
ft.-lb.
57. Working Formulas — The bending moments and shears
for a number of simple cases of common occurrence are given
below. In general the bending moment and the shear vary from
point to point along a beam, and they may be conveniently repre-
sented on a diagram by ordinates whose lengths represent the
values of those quantities. In the accompanying figures the
upper diagram is one of bending moments and the lower is one
of shears. Positive values are laid off above the base line and
negative below. The student who has followed the examples
of the preceding sections should have no difficulty in computing
the ordinates given in Figs. 12 to 15.
Figs. 12 and 13 represent cantilever beams, one carrying a
load of W at the end, the other carrying a uniformly distributed
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BEAMS.
47
load of w per unit of length. The bending moment at any
section distant x from the free end of the beam of Fig. 13
is Af= —wx-$x= — \wx 2 and the bending-moment diagram is
therefore a parabola.
w
**?
Fig. 12
.irf
Fig. 14
Fig. 13
Fig. 14 shows a beam on two supports carrying a load W.
The bending moment is evidently a maximum under the weight,
and is equal to W — j — , a quantity easily remembered as the
weight into the product of the two segments divided by the span.
When the load is at mid-span M = \WL In Fig. 15 a load of
intensity w per unit of length is distributed over the beam. The
left reaction is \wl, and the bending moment at a section dis-
tant x from the left support is
M = \wl • x — wx' \x = \wx{l — x) .
The right-hand member of this equation contains x(l— x), which
is the product of two variables whose sum is constant* therefore
the bending-moment diagram is a parabola.
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4 8
STRUCTURAL MECHANICS.
The above diagrams are drawn for the applied loads alone,
that is, the beams are considered to be without weight. If a
beam weighing w per unit of length carries a load W as in Fig.
12 or 14, the bending moment or shear at any section can be
found by adding the ordinates of Figs. 12 and 13 or of Figs. 14
and 15.
Fig. 17
At any section distant x from one end of a beam on two sup-
ports, the bending moment due to a single moving load is a
maximum when the load is at the section, when it has a value of
M = Wx{l-x)+l. As this is equation of a parabola of altitude
\Wl, the absolute maximum moments which can occur in the beam
under a single moving load are as shown in Fig. 16. The greatest
positive shear at any section occurs when the load is just to the
right of that section; it is equal to the left reaction and consequently
is proportional to the distance of that section from the right
support, hence the locus of maximum positive shear is a straight
line as shown.
If a uniform load advances continuously from the right end
of the beam of Fig. 17, the positive shear at any section will
increase until the load reaches the section, after which it will
decrease as the load extends into the left segment. This is evident
from the fact that any load placed in the right segment causes
positive shear at the section, while any load in the left segment
causes negative shear at the section; see Fig. 14. When the
load extends a distance, x, from the right support the shear at
the head of the load is F=P\ = \wx?+l and the curve of max-
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BEAMS.
49
imum shears is a parabola. To cause maximum negative shears
the load must advance from the left. The maximum moment at
any section occurs when the whole span is loaded, in which case
the moments are given by Fig. 15.
58. Position of Wheel Concentrations for Maximum Moment
at any Given Point. — Where moving loads have definite magni-
tudes and spacings, as is the case with the wheel weights of a
locomotive, the position of the load, on a beam or girder supported
at both ends, to give maximum bending moment at any given
A
*1
-*T-
£
.f
nnOOOO no on
C
Fig. 18
H
section may be found as follows : — Let the given section be C, at
a distance a from the left abutment of a beam A B, of span /,
Fig. 18.
Let -Ri be the resultant of all loads on the left of C, and acting
at a distance X\ from the left abutment; let R2 be the resultant
of all loads to the right of C, and acting at a distance x 2 from the
right abutment. The reaction Pi at A, due to R2 alone, is
R2X2 +1, and the bending moment at C, due to R 2 only, is R 2 ax 2 -*-/-
Similarly P 2 , due to R\, is R1X1 -^/, and the bending moment at C,
due to R\ only, is R\(l —d)xi -*-/. Hence the total bending moment
at C
is
M=R 2 ^ +Rl tf±.
If the entire system of loads is advanced a short distance d
to the left, the bending moment at C becomes
w t> <*(*2+<*) , p (l-a)(x t -d)
M'=K2 1 +K\ 1 *-.
The change of bending moment due to moving the loads to the
left is
ad (l—a)d
M'-M=R 2 j-R l ^-j- L .
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SO STRUCTURAL MECHANICS.
If the loads are moved a distance, d, to the right instead of
to the left, the change of bending moment is
ad (l—d)d
M'-M= -R 2T +R^—j^.
From the two values of M' —M it is seen that the bending
moment at C will be increased by moving the loads to the
left when R2fl>>R\{l—a) and by moving to the right when
Ri(l—a)>R2fl- When Ri(l— a)=i?20 the bending moment can-
not be increased by moving either to the right or left and is there-
fore a maximum. The condition may be written
R\ i?2 R\ -fi?2
a ~l—a~~ I
Ri +a is the average load per unit of length on the left segment
and (R\ +R2) -*-/ is the average load on the span, hence
The bending moment at any point of a beam carrying a system
of moving loads is a maximum when the average load on one
segment is equal to the average load on the span.
Since, for maximum bending moment at any section, a load
must be at that section, place a load W n at the given point and
compute the above inequality, first considering W n as being just
to the right and then just to the left of the section. If the inequality
changes sign, the position with W n at the section is one of M max.
The value of M max. can be computed as in § 53. If, however,
the inequality does not change sign, move the whole system until
the next W comes to the section, and test the inequality again.
It sometimes happens that two or more different positions of
the load will satisfy the condition just explained, and, to determine
the absolute M max., each must be worked out numerically.
When there are some W's much heavier than others, M max. is
likely to occur under some one of them. When other loads are
brought on at the right, or pass off at the left, they must not be
overlooked.
59. Position of Wheel Concentrations for Maximum Shear
at any Given Point. — The shear at point C in the beam or girder
of Fig. 19, as the load comes on at the right end, will increase
until the first wheel W\ reaches C. When that wheel passes C,
the shear at that point 'suddenly diminishes by W\, and then
again gradually increases, until W2 reaches C. Let R be the sum*
of all loads on the span when W\ is at C, and x the distance from
the centre of gravity of the loads to the right point of cupport
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BEAMS.
5*
B. The shear at C will be P\=Rx+L If the train moves to
the left a distance b, the space between W\ and W 2 , so that
W 2 has just reached C, the shear at C will be R(x+b) +l—Wu
plus a small quantity k which is the increase in Pi, due to
any additional loads which may have come on the span during
this advance of the train. The shear at C will therefore be
] th ±*± )r~
Q n o
«
N't
Rio?
k'l
Q n n
\
* 1? \ ■
Fig. 19
increased by moving up W 2 , if Rb +1 +k > Wi, or (as k can often be
neglected) if
J rxr R Wi
R 1 >W 1 or 7>x .
Hence move up the next load when the average load per foot
on the span is greater than the load on the left divided by the
distance between W\ and W 2 .
Similarly, W$ should be moved to C if —j->W 2 or -r-> — 2 ,
If if c
R f being the sum of the loads on the span when W 2 is at C, and
c the distance between W 2 and W3.
It is not necessary to take account of k unless the two sides
of the inequality are nearly equal.
Example. — Span 60 ft., weights in units of 1,000 lb.
123456789 10
Weights=8 15 15 15 15 9 9 9 9 8 15
Spacing =8' t 6' 4 J' aY Y 5' 6' 5' 8' 8'
To apply test for M max. at 15 ft. from left, load advancing from
right. With W 2 at quarter span, load on span =104. If W 2 is just
to the right, J -£>^ or — 4 >-; if W 2 is just to the left, ^>^;
60 15 41 41*
therefore move up W z to right of quarter span. W 10 now is on the
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52 STRUCTURAL MECHANICS.
112 23 112 ^8
span. >— • Consider W 3 to be just to the left; then — <— >
41 4 1
or the inequality changes with W3.
-Pi = (8- 59+ 15' 5 I + 45'4oH36- 21+8-5) ^60=64.26.
M ma#.= (64.26)15 — 8-14— 15-6=761,900 ft.-lb.
To test for F max. at same point. Put W\ at quarter span. Load
on span=95. |* > -. Move up W 2 \ load on span now 104. — <-^.
OO o OO O
Inequality changes. Pi = (8-53+i5-45+ 45•34H3 6 • I 5)- f - 6o== 53• 2 •
F. max.=Pi — ^1 = 53.2— 8=45,200 lb.
When these locomotive wheel loads are distributed to the
panel-joints of a bridge truss through the longitudinal stringers,
which span the panel distance between floor-beams, the above
rule is modified. A load in a panel being supported directly
by the stringers is by that means carried to the joints of the truss.
When the train advances from the right end until the forward
wheels are in the panel under investigation, the shear in that panel
is the left reaction mihus such part of the loads in the panel as
go to the floor-beam to the left. Let R be the resultant of all
the loads on the span applied at a distance x from the right sup-
port, and let R\ be the resultant of the loads in the panel, of length
p, applied at a distance, X\, from the right floor-beam. Then
I p
If the loads are moved a distance, d, to the left, the change of
shear is
d d
F'-F=R T -Ri->
I p
and by an argument similar to that of the preceding section, for
a maximum
RRi
I- p-
Hence the shear in any panel of a truss is a maximum when
the average load in the panel is equal to the average load on the
span.
60. Absolute Maximum Bending Moment on a Beam under
Moving Loads. — When a beam or girder of uniform cross-section,
such as a rolled I beam, supported at its ends, is subjected to
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BEAMS. 53
a system of passing loads, such as an engine, heavy truck, or
trolley, it generally suffices to determine that position of the
system of weights which causes the absolute maximum bending
moment, the section where it is found, and its amount.
In Fig. 18 let C be that section. Let R = resultant of all
loads on the beam and #=its distance from B; 2?i = resultant
of the loads to the left of C. The reaction at left, P x =Rx+l;
and, since the bending moment at C is to be a maximum, the
shear at C must be zero, or
Rj-R x =o. /. 7=—.
But the position of loads must also satisfy the condition of
§ 58, since there is to be maximum bending moment at C, and
R\ R R\ R\
— =-7-. /. — = — , or a=x.
a I ax
The point of absolute maximum bending moment therefore
is as far from one end as the centre of gravity of the whole load
is from the other. This rule may be written: When the middle
0} the span bisects the distance between the centre 0} gravity 0} the
whole load on the span and one of the wheels on either side of
the centre of gravity, the desired moment is to be found under
one of the two wheels.
Example. — Beam of span 24 ft. Two wheels 6 ft. apart, one
carrying 2,000 lb. and one 4,000 lb., pass across. Centre of gravity
is 2,000-6-4-6,000=2 ft. from the heavier wheel. Then this wheel is
to be placed 1 ft. from mid-span. React ion = 6,000-11-^-24 =2, 7 50
lb. M. *»a#.= 2,75o- 11 = 30,250 ft.-lb.
61. Total Tension Equals Total Compression. — If a beam,
loaded in any manner, and in equilibrium under the moments
caused by the external forces, is cut perpendicularly across by
an imaginary plane of section, while the right-handed and left-
handed bending moments already shown to exist, § 53, continue
to act, it is evident that the left and right segments of the beam
can only be restrained from revolving about this section by the
internal stresses exerted between the material particles con-
tiguous to the section. These stresses must be of such signs,
that is tensile and compressive; of such magnitude, provided
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54 STRUCTURAL MECHANICS.
the material does not give way; and so distributed over the
cross-section, as to make a resisting moment just equal to the
bending moment at the section. For the former is caused by
the latter and balances it.
Since the moment arms of these stresses lie in the perpen-
dicular plane of section, the components to be considered now
will be normal to the section. The tangential components are
caused by and balance the external shear.
As the external forces which tend to bend a beam are all
transverse to it, and have no horizontal components, the internal
stresses of tension and compression which are caused by the
bending moment must be equal and opposite, as required for
a moment or couple, and hence the total normal internal tension
on any section must equal the total normal compression.
When any oblique or longitudinal external forces act on
a beam, there is always found that resultant normal stress on
any right section which is required to give equilibrium.
62. Distribution of Internal Stress on any Cross-section. —
It may be convenient in the beginning to consider one segment
of the beam removed, and equilibrium to be assured between
the external moment tending to rotate the remaining segment
and the resisting moment developed in the beam at the section,
as shown in Fig. 20.
If two parallel lines near together are drawn on the side of
a beam, perpendicularly to its length, before it is loaded, these
lines, when the beam is loaded to any reasonable amount and
bent by that loading, will still be straight, as far as can be
observed from most careful examination; but they will now con-
verge to a point known as the centre of curvature for that part
of the beam. •
An assumption, then, that any and all right sections of the
beam, being plane before flexure, are still plane after the flexure
of this beam, is reasonable. If the right sections become warped,
that warping would apparently cause a cumulative endwise
movement of the particles at successive sections, especially in a
beam subjected to a constant maximum bending moment over a
considerable portion of its span; and such a movement and
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BEAMS.
55
resulting distortion of the trace of the sectional plane ought there-
fore to become apparent to the eye. Such a warping can be
perceived in shafts, other than cylindrical, subjected to a twisting
couple, but cannot be found in beams.
The lines A C and B D just referred to will be found to be
farther apart at the convex side of the beam, and nearer together
at the concave side, than they first were; hence a linfe G H, lying
somewhere between A B and C D, is unchanged in length. If,
in Fig. 20, a line parallel to A C is drawn through H, the extremity
ft c ED FIg.»
of the fibre GH which has not changed in length, KL will
represent the shortening which I L has undergone in its reduction
to I K, and N O will represent the lengthening which M N has
experienced in stretching to M O. The lengthening or shorten-
ing of the fibres, whose length was originally G H=ds, is directly
proportional to the distance of the fibre from G H, the place of
no change of length, and hence of no longitudinal or normal
stress.
The diagram, Fig. 1, representing the elongation or shortening
of a bar under increasing stresses, shows that, for stresses within
the elastic limit, equal increments of lengthening and shortening
are occasioned by equal increments of stress. If this beam has
not been loaded so heavily as to produce a unit stress on any
particle in excess of the elastic limit (and no working beam, one
expected to last permanently, should be loaded to excess), the
longitudinal unit stresses between the particles will vary as the
lengthening and shortening of these fibres, that is, as the dis-
tance from the point of no stress. Hence, at any section, the
direct stress is uniformly varying, with a maximum tension on
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STRUCTURAL MECHANICS.
the convex side and a maximum compression on the concave
side.
The stresses on different forms of cross-section A C are shown
in Fig. 21. The total tension on the section is always equal to
the total compression.
Fig. 21
63. Neutral Axis. — The arrows in Figs. 20, 21 may be taken
to represent the unit stress at each point of the cross-section,
varying as the distance from the plane of no stress, and constant
in the direction z. To locate the point or plane of no stress or
neutral axis for successive sections :
Let } c and } t be the unit stresses of compression and tension
between the particles at the extreme edge of any section, distant
y c and y t from the point of no stress. It is plain that fe'ft—ye'yt
from similar triangles, and that the unit stress p at any point
distant y from the point of no stress will be
r y/
h
or -y,
yt
or, in general — y,
from a similar proportion.
If zdy is the area of the strip on which the unit stress p is
exerted, z being the variable coordinate at right angles to x and
y, the total force on zdy will be pzdy = — yzdy } where — is a con-
stant, the unit stress at a unit distance.
As the total normal tension on the section is to equal the
total compression, or their sum is to be zero, § 6i, the condition
may be written
I r+y*
y\J -yt
y\
Therefore the sum of the moments zdy -y of the strips zdy about
the axis of z must balance or be zero. Then the axis of z or
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BEAMS. 57
neutral axis must pass through the centre of gravity of a thin plate
representing the section, and the neutral axis of any section lies
in its plane, and usually in a direction perpendicular to the plane
oj the applied external forces. The axes of the successive cross-
sections make up what is known as the neutral plane of the beam.
Although there is no longitudinal or normal tension or compres-
sion at that line of the cross-section, it experiences shear, as will
be shown later.
64. Resisting Moment. — The law of the variation of stress
on the cross-section and the location of the neutral axis have been
established. The resisting moment is caused by and is equal to
the bending moment. The moments of all the stresses about the
neutral axis Z Z is, since p has the same sign as y, and the moments
conspire,
M = f(+p)zdy(+y)+f(-p)zdy(-y) = f_°pyzdy.
As }+yi denotes the unit stress at either extreme fibre divided
by its distance from the neutral axis, and p= —y,
M
=yj-y t yzdy= n' and ; " '
where / represents fy^zdy about the axis Z Z, lying in the plane
of the section, through the centre of gravity of the same and
perpendicular to the plane of the external forces applied to the
beam. / is termed in mechanics the moment oj inertia of a plane
area, and is usually one of the principal moments of inertia of
the area. The integral will be of the fourth power, involving
the breadth and the cube of the depth. For moments of inertia
of plane sections, see Chap. IV.
In the above expression for the resisting moment the quantity
I+yi is known as the section modulus. The section moduli of
steel beams, angles, etc., are tabulated in the handbooks published
by the various steel manufacturers, so that the resisting moment
of a steel beam can be readily found by multiplying the section
modulus by the working stress.
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53 STRUCTURAL MECHANICS.
•
As moments of inertia of plane areas are of the fourth power,
and can be represented by n'ftA 3 , where h is the extreme dimension
parallel to y, and b to z, and as y\ may be written m'A, the resisting
moment can be represented, if n'-7-m'=», by
yi
n being a fraction. For a rectangular section this becomes
1 12 2 6' '
and for a circular section
M = / •£- -5- -d = — /d 3 = 0.0982/d 3 .
Examples. — A timber beam 6"Xi2", set on edge, with a safe unit
stress of 800 lb. will safely resist a bending moment amounting to
8oo-6-i2 2 ^6= 115,200 in.-lb.
A round shaft, 3 in. in diameter, if /= 12,000 lb. will have a safe
resisting moment of 12,000- 22 -3 s -s-7 -32 =31,820 in.-lb.
For rectangular sections, either b or h is usually assumed and
A or ft then found. If the ratio h+l is fixed by the desire to
secure a certain degree of stiffness (see "Deflection of Beams,"
Chap. VI.) 1 the unknown quantity is b.
Example. — A wooden beam of 12 ft. span carries 3,600 lb. uniformly
distributed. M=JW7=J-3,6oo- 12- 12=64,800 in.-lb. If /=i,ooo,
£=1,400,000, and the deflection v is not to exceed jfa of the span,
v «;//.,., 1 5 1,000-2-144 , . ___
from T = -^r- is obtained ;— = -%• - n? ; .'. h= 13 in. Then
/ 48^1 600 48 1,400,000- h °
assuming ^=14 in., a practicable size, 64,800= -Z-r— b> 14 2 ; and 6=
2 in.
Economy of material apparently calls for as large a value of
h as possible; but the breadth b must be sufficient to give lateral
stiffness to the beam, or it may fail by the buckling or sidewise
flexure of the compression edge, between those points where it
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BEAMS. 59
is stayed laterally. The effect of loading as a beam a thin
board set on edge will make clear the tendency.
When the plane of the applied forces does not pass through
the axis of the beam, a twisting or torsional moment is added,
which will be discussed in § 86.
65. Limit of Application of M=f I-J-yi. — The expression for
the resisting moment at any section of a beam, caused by and
always equal to the external bending moment at that section, is
applicable only when the maximum unit stress / does not exceed
the unit stress at the elastic limit of the material. If / exceeds
that limit, a uniformly varying stress over the whole section is
not found, and the neutral axis may not remain at the centre
of gravity. Hence, also, the substitution of breaking weights,
obtained by experiments on beams which fail, in a bending-
moment formula which is then equated with fl+yu results in
values of /, the then so-called modulus 0} rupture, agreeing with
neither the tensile nor the compressive strength of the material,
and therefore of but limited value. This formula is correct for
the purpose of design and construction; but its limitation should
be kept in mind.
66. The Smaller Value of f-^yi to be Used. — Since from
similar triangles fc+yc^h+Vt* *t ls immaterial which ratio is
used for M for a given cross-section. But, in designing a cross-
section to resist a given moment, if y t and y c are not to be equal,
another consideration has weight. A numerical example will
bring out the distinction.
A beam of 24 in. span is loaded at the middle with a weight
of 500 lb. M max. will be \Wl= 500 6 = 3,000 in.-lb. If the
depth of the beam is 5 in., and its section is of such a form that
the distance from its centre of gravity to the lower edge is 2 in.,
and to the upper edge is 3 in., while 7=4, then 3,000 = £/<• 4
or J/ c • 4. Hence the maximum unit tension j t = 1,500 lb. per sq. in.,
and the maximum unit compression / c = 2,250 lb. per sq. in.
But if the material of the above beam must not be subjected
to a unit stress greater than 2,000 lb. per sq. in., that unit stress
will be found on the compression side; for 2,000 lb. per sq. in.
on the tension side would be accompanied by 3,000 lb. per sq.
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60 STRUCTURAL MECHANICS.
in. on the compression side; and a unit stress of 2,000 lb. com-
pression is only compatible in this case with 2,000 § = 1,333 lb.
unit stress tension. The beam will safely carry only a moment
of 2,000-4^-3 = 2,667 in.-lb.
Hence, when designing, with a maximum allowed value of
/, and using a form of section where y t and y c differ, take that
ratio of /-*-yi which is the smaller. For a few materials, where
j e and jt may be taken as differing in magnitude, as perhaps in
cast iron, use that ratio } c ■ i ry c or j t +y t which gives the smaller
value. As the elastic limit in tension and compression for a
given material is usually the same, use in computations the larger
value of y\.
67. Curved Beams. — An originally curved beam, at any
given cross-section made at right angles to its neutral axis, so
far as the resisting stresses to bending moments are concerned,
is in the same condition with an originally straight beam at a
similar and equal cross-section to which the same bending moment
is applied. Any definite thrust or tension at its two ends adds
a moment at each right section equal to the product of the force
into the perpendicular ordinate from the chord to the centre of
the section, and a force, parallel to the chord, which force can
be resolved into one normal to the section and a shear. Compare
Fig. 11.
68. Inclined Beams. — A sloping beam is to be treated like
a horizontal beam, so far as resisting stress produced by that
component of the load which is normal to the beam is concerned.
The component of the load which acts along the beam is to be
considered as producing a direct thrust along the beam if taken
up at the lower end; or a direct tension, if taken up at the upper
end, or as divided somewhat indeterminately, if resisted at both
ends. If this longitudinal force is axial, the mean unit stress j c
caused by it is to be added to the stress }b of the same kind from
bending moment at the section where this sum j e +/& will be a
maximum. This point can easily be found graphically. If
the section of the piece is the unknown quantity, it will commonly
suffice to use the value of M max. to determine an approximation
to j bl and to correct the section by the resulting value of J c +Jb
at the point where the sum is largest.
If the direct force at the end or ends is not applied axially,
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BEAMS. 6 1
its moment at any section may augment or diminish the bending
moment of the normal components of the load.
Cases of inclined beams, for a given load and inclination, are
better solved directly than by the application of formulas.
Example. — A wooden rafter, 15 ft. long, has a horizontal pro-
jection of 12 ft., and a rise of 9 ft., and it carries a uniformly distributed
load of 1,500 lb. The normal component of this load will be 1,200 lb.,
the component along the roof 900 lb. The maximum bending moment,
1 .tn mi t I,200Xl5Xl2 . 11 -rr , e
at the middle, will be r-^ = 27,000 m.-lb. If the safe
o
11 2
stress is 1,000 lb., the section to carry this moment should be — — - — =■
6
27,000, or bh 2 = 162. If £=3, h=8 in. If the mean thrust, at the
middle of the rafter, is 1,250 lb., the maximum thrust, at the bottom
end, will be 1,700 lb., and the minimum thrust, at the top end, will
be 800 lb. The section of maximum fibre stress will be a very little
below the middle. But, if the rafter is 3"X8", /& from bending
... , 27,ooo'6 _ „ A . . 1,250 ..
moment will be ' fl =844 lb. Also, / c = — — = 52 lb. Hence
3'S'S ■ ' 24 J
/c+/& = 8o6 lb., a satisfactory result, if the rafter is stayed laterally by
the roof-covering or otherwise.
69. Movement of Neutral Axis if Yield-point is Exceeded. —
If it is assumed that cross-sections of a beam still remain plane
after the yield-point is passed at the extreme fibres, the stretch
and shortening of the fibres at any cross-section will continue
to vary with the distance from the neutral axis or plane. Suppose
then that the elongation per unit of length of the outer tension
fibre has attained an amount equal to O L, Fig. 1. The unit
stress on that fibre will be L N. A fibre lying half-way from that
edge to the neutral axis will have a unit stress K M. If the beam
is rectangular, the total tension on the cross-section must be the
area OMNL,OL now being the distance from the neutral axis
of the beam to the tension edge. Since the total compression on
the section must equal the total tension, an equal area O L'N'
must be cut off by L' N' and the compression curve. The neu-
tral axis must then divide the given depth of the beam in the
ratio of O L to O L', shifting in this case towards the compression
side. Had the compression curve been below the tension curve,
the neutral axis would have shifted towards the convex side of
the beam.
Since L N is less than L' N', the unit stress on the extreme
fibre on the tension side is the less. Hence this displacement of
the neutral axis favors the weaker side. If such action continued
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62 STRUCTURAL MECHANICS.
to the time of fracture, it would account for the fact that the
application of the usual formula, fl+yu to breaking moments
gives a value of / which lies between the ultimate tensile and
compressive strengths of the material. It must be borne in mind,
however, that the compression portion of the section increases
in breadth and the tension portion contracts, quite materially
for ductile substances, thus adding to the complication. A
soft steel bar cannot be broken by flexure as a beam at a i ingle
test.
A rectangular cross-section also tends to assume the section
shown in Fig. 22. The compressed particles in the middle of the
width can move up more readily than they can
laterally, making the upper surface convex as well
as wider, and the particles below at the edges,
being drawn or forced in, are crowded down,
making the lower surface concave as well as
narrower.
'^ig.af 1 Hence the position of the neutral axis is un-
certain, after the yield-point has been passed on
either face; but it is probably moved towards the stronger side.
70. Cross-section of Equal Strength. — When a material will
safely resist greater compression than tension, or the reverse,
it is sometimes the custom to use such a form of cross-section
that the centre of gravity lies nearer the weaker side. Ca£t
iron is properly used in sections of this sort. See Fig. 21, section
at right. Wrought-iron or steel sections are occasionally rolled
or built up in a similar fashion, but the increase in width of the
compression flange is then usually intended to increase its lateral
stiffness.
If /<=safe unit tensile stress, and / c =safe unit compressive
stress, the centre of gravity of the section must be found at
such point that y*:y c =/<:/« when the given safe stresses will
occur simultaneously at the section. By composition, yt>y c '-h
=*U'-U'U +/« so * at ^ e centre of gravity should be distant
from the bottom or top,
*-*&+/? or ye=h m
Example. — If /<=3,ooo lb., and / c =9,ooo lb., y t = h -3,000-?- 12,000
■■ \h. If a cast-iron £ section is to be used, base 10 in., thickness through-
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BEAMS. 63
out of i in., and height of web h\ then, by moments around base,
i( * + l) — IO +y ; •"' * S m '' ^ =BI * m,;
_ 10 , 125, , 165 -. 3,000* 165-2 _ . „
7= — h— ? + io-i + s-4=— -. M=°- 3 — =82,500 in.-lb.,
12 12 ° 4 4-3
the moment that the section will carry.
71. Beam of Uniform Strength. — As has been shown in
§ 64, the resisting moment may be put into the form M =njbh 2 ,
where n is a numerical factor depending on the form of cross-
section. If, then, for a given load, bh 2 be varied at successive
cross-sections to correspond with the variation of the external
bending moment, the unit stress on the extreme fibre will be
constant; the beam will be equally strong at all sections, except
against shear; and there will be no waste of material for a given
type of cross-section, provided material is not wasted in shaping.
Suppose, for example, that a beam is to be supported at
its ends, to carry W at the middle, and to be rectangular in cross-
section. The bending moment at any point between one sup-
port and the middle is \Wx. Equate this value with the resist-
ing moment. \Wx=\jbh 2 . To make / constant at all cross-
sections, bh 2 must vary as x from each end to the middle. If h
is constant, b must vary as x, or the beam will be lozenge-shaped
in plan and rectangular in elevation. If, on the other hand,
b is constant, h 2 must vary as x, and the elevation will consist
of two parabolas with vertices at the ends of the beam and axis
horizontal, while the plan will be rectangular.
The section need not be a rectangle. If the ratio of b to
h is not fixed, the treatment will be like the above; but, if that
ratio is fixed, as for a circular section, or other regular figure,
b=ch, and ft 3 must vary as the external bending moment, or,
in the case above, as x. The cross-section of the cast-iron beam
in the example of the previous section may be varied in accordance
with these principles.
The following table gives the shape of beams of rectangular
cross-section supported and loaded as stated.
When a beam supported at both ends carries a single moving
load W 9 passing across the beam, the bending moment at the
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64
STRUCTURAL MECHANICS.
Beam.
M.
bh*
varies a
h* Constant,
b varies as
b Constant,
h a varies as
Fixed at one end,
W at other.
Fixed at one end,
uniform load.
SupYdbothends
W at a from
end.
SupYd both ends,
uniform load.
-Wx
%wx(l—x)
X
X*
X
l-x
x, triangular plan,
Fig. 23.
x 2 , parabolic plan,
Fig. 25.
x 1 triangular plan,
/-J Rg - 27 -
x(t—x) parabolic plan.
Fig. 29.
x, parabolic eleva-
tion. Fig. 24.
x*, h varies as x,
triangular eleva-
tion. Fig. 26.
x 1 parabolic ete-
[ vation.
/_J Fig. 28.
x(l—x1, circular
or elliptical eleva-
tion. Fig. 30.
point x, where the load is at any instant, = Wx(l — x) h-/. Such
a beam will therefore fall under the last class of the above table.
Fig. 23
Fig. 27
Fig. 24
Fig. 28
Fig. 25
Fig. 29
Fig. 26
Fig. 80
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BEAMS. 65
Beams which can be cast in form or built up may be made
in the above outlines, if desired. Some common examples,
such as brackets, girders of varying depth, walking-beams, cranks,
grate-bars, etc., are more or less close approximations to such
forms. Enough material must also be found at any section
to resist the shear, as at the ends of beams supported at the ends.
Where a plate girder is used (see Fig. 95) with a constant
depth, the cross-section of the flanges, or their thickness when
their breadth is constant, will theoretically and approximately
follow the fourth column of the preceding table. If the flange
section is to be constant or nearly so, the depth must vary in the
same way, and not as in the fifth column.
Roof- and bridge-trusses are beams of approximate uniform
strength, for the different allowable unit stresses and for chang-
ing loads. The principles of this section have an influence on
the choice of outline for such trusses, and the shapes of moment
diagrams suggest truss forms.
72. Distribution of Shearing Stress in the Section of a Beam,
Pin, etc. — It will be proved, in § 154, that, at any point in a
body under stress, the unit shear on a pair of planes at right
angles must be equal. Whatever can be proved true in regard
to the unit shear on a longitudinal plane at any point in a beam
must therefore be true of the unit shear on a transverse plane at
the same point.
Fig. 31 represents a portion of a beam bent under any load.
The existence of shear on planes parallel to E F is shown by the
tendency of the layers to slide by one
another upon flexure. Let the cross-
section of the beam be constant. If
the bending moment at section H, a/
point close to G, differs from that at G, F} n
there will be a shear on the transverse
section, because the shear is the first derivative of the bending
moment, § 56. The direct stress, here compression, on the
face H F of the solid H F E G, will differ from that on the face
G E, since the bending moments are different, and that difference
will be balanced by a longitudinal horizontal force,- or shear,
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66 STRUCTURAL MECHANICS.
on the plane FE, to oppose the tendency to displacement. If
this force along the plane E F is divided by the area E F over
which it is distributed, the longitudinal unit shear will be obtained.
It follows from the first paragraph that the unit shear at the
point E on the transverse section G A must be the same. It is
also evident that the farther E F is taken from H G, the greater
will be the difference between the total force on H F and that
on G E, until the neutral axis is reached, and that the unit shear
on the longitudinal plane E F must increase as E F approaches
B, the neutral axis. The same thing is true if the plane is sup-
posed to lie at different distances from the edge A. Hence, at
any transverse section A G, the unit shear on a longitudinal plane
is most intense at the neutral axis ; and therefore the unit shear
on a transverse section A G is unequally distributed, being greatect
at B, the neutral axis, and diminishing to zero at A and G.
Pins and keys, and rivets which do not fit tightly in their
holes, and hence are exposed to bending, have a maximum unit
shear at the centre of any cross-section, and this shear muLt
therefore be greater than the mean value, and must determine
the necessary section.
To find the mathematical expression for the variation of
shear on the plane A G:
O B D C is the trace of the neutral plane. BD = EF sensibly
=dx. B E=y, B G=yi. Breadth of beam at any point = z, at
neutral axis=3o. Normal or direct unit stress at the point E
on plane AG = />. Unit shear at E = q; maximum, at B, = j .
M and F= bending moment and shearing force at section A G.
By §64, /-^ and p=jy.
The total direct stress on plane G E is
J y p^y-TJy yzdy - < x)
The difference between M at the section through B and M
at the section through D must be Fdx, since M=fFdx, by § 56.
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BEAMS. 67
The horizontal force on E F is the excess of (1) for G E over
Fdx p*
its value for H F, or -j— / yzdy. Divide by the area zjix
of F E, over which this horizontal force acts, to find the unit shear.
Since the mean unit shear =jF-^S, the ratio of the maximum
unit shear to the mean will be found by dividing jo by F+S,
Max. unit shear S P*
-7z « y* d y=
f^yzdy
r2*„ »
Mean unit shear IzqJo j j t 2 zq
where r— radius of gyration of the cross-section, and J yzdy
is the moment of either the upper or lower part of the cross-
section about the trace of the neutral plane. Hence the max-
imum unit shear will be:
bJ Q ydy i2h2
Rectangle, 1^2.5 = "Ws = P 0T 5 °'° 8 reater *
c,rcl Vo W?R = ~W? or 33% 8reater -
Thin ring approximately = 2, or 100 per cent, greater than the
mean unit shear.
For beams of variable cross- section / will not be constant;
but the preceding results are near enough the truth for prac-
tical purposes.
Example. — A cylindrical bridge-pin 3 in. diam., area 7.07 sq. in.,
has a shear of 50,000 lb. The maximum unit shear is — =
70.7 3
9,430 lb. per sq. in.
If the apparent allowable unit shear is reduced one quarter,
as from 10,000 lb. to 7,500 lb., the same circular section for a
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68 STRUCTURAL MECHANICS.
pin will be obtained in designing as if the maximum unit shear
were considered. For a rectangular section the apparent allow-
able shear should be reduced one-third.
Example. — A 4"X6" beam has at a certain section a shear of 2,400
lb. ; the maximum unit shear on both the horizontal and the vertical
plane, at the middle of the depth, is ■ '—=150 lb. per sq. in.
24 2
As the shearing resistance along the grain of timber is much
less than the shearing resistance across the grain, wooden beams
which fail by shearing fracture along the grain at or near the
neutral axis, at that section where the external shear is greatest
As the unit shears on two planes at right angles through a given
point are always equal, the shearing strength of timber across
the grain cannot be availed of, since the piece will always shear
along the grain.
A rectangular timber beam of span /, carrying w per unit
of length, has a maximum fibre stress of
1 4bh 2 '
and the greatest shearing stress along the neutral axis is
Dividing the first equation by the second gives the ratio between
the maximum fibre stress and the maximum shear existing in
the beam. If the corresponding ratio between the allowable
stresses for the beam is greater than the ratio between the existing
stresses, the beam is weaker in shear than in flexure; hence if /
and q are working unit stresses, a rectangular beam carrying a
uniformly distributed load should be designed for shear when
L<1
h=qo
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BEAMS.
69
A beam carrying a single load at the centre should be designed
for shear when
l -<-L.
h — 2 jo*
73. Variation of Unit Shear.— The distribution of shear
on three forms of cross-section is indicated in Fig. 32, where
r
Pig. 32
the ordinates show the unit shear at corresponding points. For
both the rectangle and the circle the intensity of shear varies
according to the ordinates of a parabola. The curve for the
I-shaped section is made up of three parabolas as shown, the
intensity of shear being nearly constant over the web. For
I beams of ordinary proportions and for plate girders the curve
showing unit shears in the web will be much flatter than in this
figure, and the maximum unit shear will differ but little from
the unit shear found by dividing the total shear at a section by
the cross-section of the web, as is usually done in practice.
Examples. — 1. Three men carry a uniform timber 30 ft. long.
One man holds one end of the timber; the other two support the
beam on a handspike between them. Place the handspike so that
each of the two shall carry J of the weight.
2. Three sections of water-pipe, each 12 ft. long, are leaded end
to end. In lowering them into the trench, where shall the two slings
be placed so that the joints will not be strained? Neglect the extra
weight of socket.
3. Wooden floor-joists of 14 ft. span and spaced 12 in. from centre
to centre are expected to carry a floor load of 80 lb. per sq. ft. If
f—goo lb., what is a suitable size? 2"X 10".
4. One of these joists comes at the side of an opening 4'X6', the
load from the shorter joists, then 10 ft. long, being brought on this
longer joist at 4 ft. from one end. How thick should this joist be ?
4 in.
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7© STRUCTURAL MECHANICS.
5. A cylindrical water-tank, radius 20 ft., is supported on I beams
radiating from the centre. These beams are supported at one end
under the centre of the tank and also on a circular girder of 15 ft.
radius. They are spaced 3 ft. apart at their outer extremities. If
the load is 2,000 lb. per sq. ft. of bottom of tank, find the max.-f and
—if on a beam. +29,600; —68,750 ft.-lb.
6. Find b and h for the strongest rectangular beam that can be
sawed from a round log of diameter d.
7. A opening 10 ft. wide, in a 16-in. brick wall, is spanned by a
beam supported at its ends. The maxiipum load will be a triangle
of brick 3 ft. high at the mid-span. If the brickwork weighs 112 lb.
per cu. ft., find M at the mid-span. 44,800 in.-lb.
8. In the above problem, write an expression for M at any point
if w= weight of unit volume of load, a = height of load at middle,
i=» thickness of wall, /=span, and #= distance of point from the sup-
port.
9. A trolley weighing 2,000 lb. runs across a beam 6 in. wide and
of 20 ft. span. What will be the elevation of a beam of uniform strength,
and what its depth at middle, if /=8oo lb. ? 12 in.-f
10. A round steel pin is acted upon by two forces perpendicular
to its axis, a thrust of 3,000 lb. applied at 8 in. from the fixed end of
the pin, and a pull of 2,000 lb. applied 6 in. from the fixed end and
making an angle of 6o° with the direction of the first force. Find
the size of the pin, if /= 8,000 lb. If =20,784 in.-lb.
11. A beam of 20 ft. span carries two wheels 6 ft. apart longitudi-
nally, and weighing 8,000 lb. each. When they pass across the span,
where and what is M max. ? 57,8oo ft.-lb.
12. A floor-beam for a bridge spans the roadway a and projects
under each sidewalk b. If dead load per foot is w, live load for road-
way «/, for sidewalk «/', write expressions for +M max. and — M max.
13. A vessel is 200 ft. long. It carries 5 tons per ft. uniformly
distributed, and a central load of 300 tons. Find M max. when at
rest; when supported on a wave crest at bow and stern with each
bearing 20 ft. long; and when supported amidships only with bearing
30 ft. long.
14. The end of a beam 6 in. wide is built into a wall 18 in. The
bending moment at the wall is 600,000 in.-lb. If the top of the beam
bears for 9 in. with a uniformly varying pressure and the bottom the
same, what is the maximum unit compression on the bearing surface ?
1,852 lb.
X
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CHAPTER IV.
MOMENTS OF INERTIA OF PLANE AREAS
74. Definitions. — The rectangular moment 0} inertia, J, of
a plane area about an axis lying in that plane is the sum of the
products of each elementary area into the square of its distance
from the axis. The plane areas whose moments of inertia are
sought are commonly cross-sections of beams, and unless other-
wise stated the axis about which moments are taken passes
through the centre of gravity of the cross-section. The quotient
of the moment of inertia by the area is the square of the radius
oj gyration, r. If the area be referred to rectangular axes Z
and Y, and if subscripts denote the axes about which moments
are taken,
I, = fffdzdy = fzfdy=Sr*;
I y = ffz 2 dzdy = fz 2 ydz - Sr v 2 .
If each elementary area be multiplied by the square of its
distance from an axis perpendicular to the plane and passing
through the centre of gravity, the summation gives the polar,
moment of inertia, J. As the distance of each elementary area
from the axis is Vz 2 +y 2 ,
When the term "moment of inertia" is used without qualification,
the rectangular moment is meant. Moments of inertia, being
the product of an area into the square of a distance, are of the
fourth power and positive.
71
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j 2 STRUCTURAL MECHANICS.
The values of J, 7, and r 2 for some common forms of cross-
section follow.
I. Rectangle, height A, base b. Fig. 33. Axis through the
centre of gravity and parallel to b.
/+** /•+** Ti l + * h bh? bh? bh 3
v 12 12 12
For an axis through the centre of gravity and perpendicular
to the plane,
J-Ji+J.-^+tf). and f 2=^(62 +A 2).
II. Triangle, height A, base 6. Fig. 34. Axis as above and
parallel to b.
h:b= 2 -h-y.z; z-^h-y).
= AV243""324 + 243 324/ 36*
bit* bh h 2
36 ' 2 18*
' III. Isosceles triangle, about axis of symmetry. Fig. 35.
Height along axis h, base b.
h:\b~z-.\b-y; z=h(i-f).
2 ^
24
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MOMENTS OF INERTIA OF PLANE AREAS.
73
The sum of II and III will be the polar moment, /, about
an axis through the centre of gravity and perpendicular to the
plane.
12V3 4/ t>\3 4/
Fig. S3
Fig. 34
Fig. 35
Fig. 38
Fi^.37
IV. Circle, radius R, diameter d. Fig. 36. If = angle
between the axis of Z and a radius drawn to the extremity of any
element parallel to that axis,
'•-*£
y=Rsind; \z=R cos 6; dy*=R cos OdO.
7 -R* 7:d*
. T r, 17 kK* na*
sin 2 0cos 2 <W0= -2R*^\ - sin 40-0 — 7""6l'
. \xR* i i
' 2 = *—?; = -i? 2 = —TCP.
-/? 2 4 '
16
The polar moment of inertia, J, may be easily written if
r 1 — variable radius,
--I
R -R*
2
r 2 =-R 2 .
2
Since i,+/ y = /> and I y = I z by symmetry, I z = \tzR a as before.
V. Ellipse. Diameters d and b. Fig. 37.
As the value of 2 in the ellipse is to that of z in the circle,
as the respective horizontal diameters, or as b to d, and as the
moment of the strip zdy varies as the breadth alone, the ellipse
having horizontal diameter b, height d y gives
J *"6 4 '</" 64 ; f ~ 4 "16'
/y =
f ; ,.£, + „,
r 2 = _ (rf 2 +i 2).
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74 STRUCTURAL MECHANICS.
VI. The moment of inertia of a hollow section, when the
areas bounded respectively by the exterior and interior perimeters
have a common axis through their centres of gravity, can be
found by subtracting / for the latter from / for the former. Thus :
Hollow rectangle, interior dimensions V and A', exterior b
and u; I t (=-frbh*-Vh'*).
Hollow circle, interior radius R', exterior radius R;
I g = \K(R*-R'*).
ia = i(lP+fl'*)-*(iP+d'*).
The moment of inertia of a hollow ring of outside diameter
d and inside diameter d', the ratio of d! to d being n, may be
written
As the cross-section 5 = j7r(rf 2 -(/' 2 ) = ^(i— n 2 )<P,
*-,-$ and ,._f(,_£)_|(„_§.
75. Moment of Inertia about a Parallel Axis. — To find the
moment of inertia i 7 of a plane area about an axis Z parallel
to the axis Z through the centre of gravity and distant c from it.
By definition I' = f(y+ c) 2 zdy = J y 2 zdy + 2cfyzdy + c 2 Jzdy.
The first term of the second member is I t} the moment of inertia
about the axis through the centre of gravity; the second term
has for its integral the moment of the area about its centre of
gravity, which moment is zero; and the integral in the third term
is the given area S. Hence
/' = /, +c 2 S = (r 2 +c 2 )S = S 2 S.
Example. — I t for rectangle, axis parallel to b, is -kbh 3 . V about
base=rVM 3 +iA 2 -M=§M 3 , and ^ 2 =iA 2 .
The reverse process is convenient for use.
I, - /' -c 2 S - {f 2 -c 2 )5 = r 2 S.
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MOMENTS OF INERTIA OF PLANE AREAS. 75
As the value of / about an axis through the centre of gravity
h the least of all Ps about parallel axes, it can readily be seen
whether c 2 S is to be added or subtracted.
If the value of I\ about an axis distant C\ from the centre
of gravity is known, and it is desired to find I 2 about a parallel
axis distant c 2 from the centre of gravity, a combination of the
two formulas
Ii=I t +Ci 2 S and I 2 =Ig+c#S
gives
Example. — I for a triangle about an axis through the vertex parallel
to the base is easily obtained, since z'.b=*y.h.
Therefore / about vertex = / -ry z dy— — .
Jo n 4
Then 7 about base = hi — -JA 2 — = — .
4 \9 9/ 2 12
76. Moments of Inertia of Shapes. — It is frequently necessary
to divide areas, such as T, I, and built iron sections, and those
of irregular outline, into parts whose moments of inertia are
known, each about an axis through its own centre of gravity;
then to the sum of their several Ps add the sum of the products
of each smaller area into the square of the distance from its
axis to the parallel axis through the centre of gravity of the whole.
This rule is an expansion of the preceding one.
I, for the whole = II+Ic 2 S.
Formulas for such cases are of little value. In actual com-
putation follow the general rule.
Example. — Find / of a 6"X4 ,/ Xi // angle about an axis parallel to
shorter leg. Neglecting fillets as is customary, 5= 9.5X^=4.75 sq. in.
Divide figure into two parts, 6X£ s =3 sq. in. and 3.5XJ— 1.75 sq. in.
Distance of centre of gravity from middle of 6-in. leg is 3.5X0.5X 2.75 -*-
4.75=1.01 in., which is 1.99 in. from heel.
^=T 5 fXiX6 3 +3Xi.oi 2 4- 1 i F X3.5Xo.5 3 +i.75Xi.74 2 = 1740 in. 4
Or consider two rectangles, one 4"X6", the other 3$"X5$".
/=iVX4X63+ 4 X6Xi.oi 2 — r VX35X5-5 3 -35X5.5Xi.262= 17.40 m*
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76
STRUCTURAL MECHANICS.
77. Moments of Inertia for Thin Sections. — Values of J for
rolled shapes may also be approximately obtained by the follow-
ing method, and, if the values given in the manufacturers' hand-
books are not at hand, will prove serviceable.
The moment of inertia of a thin strip or
rod, Fig. 38, of length L and thickness /, about
— 4 1 an axis passing through one end of and making
an angle with it, is the same as if \tL were
concentrated at the extreme end. Let /= distance along strip
to any particle.
/ = f L tdl . P sin 2 - J/Z, 3 sin 2 = \tLy x 2 ,
or one-third the area multiplied by the square of the ordinate
to the extreme end.
This expression might be derived from J for a rectangle,
taken about one base.
If the rod is parallel to the axis, and at a distance y\ from
it, I = tL-yi 2 , since all particles are equidistant from the axis.
Example. — Find approximate value of I of angle of last example.
Using centre line, angle is 5f"X3l"X$". Distance of centre of
gravity from heel is 5.75X2.875-^9.5 = 1.74 in.
/=JXi(4.oi 3 -+-i.74 3 ) + iX 3 .75Xi.74 2 =i7.3oin*
78. Rotation of Axes. — If the moments of inertia about
two axes at right angles are known, the moments of inertia about
axes making an angle a with
the first may be found. In
the solution the quantity^ zyds
or J J zydzdy occurs, which is
called the product of inertia
and represented by Z. Prod-
ucts of inertia may be either
positive or negative according
to the position of the axes. Fiom Fig. 39,
^ = 2 cos a+ysina,
y / = y cos a— z sin a;
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MOMENTS OF INERTIA OF PLANE AREAS. 77
7/ — J (y cos a —z sin a) 2 dS
= J y 2 cos 2 adS—2jzy sin a cos adS +fz 2 sin 2 adS,
Iy = j (z cos a+y sin a) 2 </S
= j z 2 cos 2 adS +2jzy sin a cos arfS +J y 2 sin 2 ad5
Z' — J (z cos a +y sin a)(y cos a -2 sin a)</S
""/(y 2 ~ s2 ) s * n a cos a ^ +y zy cos 2 a</S -y zy sin 2 adS;
// =7, cos 2 a +/y sin 2 a — 2Z sin a cos a,
iy' = 7 y cos 2 a+I s sin 2 a + 2Z sin a cos a,
Z' = (7 t — 7 y ) sin a cos a +Z (cos 2 a —sin 2 a).
79. Principal Axes. — To find the maximum and minimum
values of // and Iy differentiate with respect to a :
dl '
-j 1 -= —2l s sin a cos a +2l y sin a cos a — 2Z cos 2 a +2Z sin 2 a,
da
£--■*■ £-•*■
Hence maximum and minimum values of 1/ and J y ' occur when
Z'=o, and since J=I t +I v =I t ' +I y ' = a, constant, J y ' is a mini-
mum when // is a maximum. The axes for which 1/ and Iy
are maximum and minimum are called principal axes. An
axis 0} symmetry is always a principal axis since for such an
axis Z must be zero, as for every positive ordinate there is a cor-
responding negative ordinate. If a figure has two axes of sym-
metry not at right angles, the moment of inertia is the same for
all axes through the centre of gravity.
The angle, <f>, which the principal axis, A A, makes with the
original Z axis may be found by making Z'=o:
(J f -I y ) sin <f> cos <f> +Z (cos 2 <£ -sin 2 0) «o,
^i^-Jy) Sin 20+2 COS 20 = 0,
tan 29= j? — , .
ly—l 9
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7*
STRUCTURAL MECHANICS.
I a, the moment of inertia about the A axis, is found by substi-
tuting the value of in the expression for 7/ :
I a =1* cos 2 <f> +I y sin 2 — Z sin 20;
cos 2 = %(i H-cos 20), sin 2 = £(i —cos 2$) ;
I& = £/,(i +cos 20) +£J y (i —cos 20) — Z sin 20
— K^V^X 1 ~ cos 2 0) "~£ s * n 2 +^ t
^ 1 -COS 20
«=Z— 7 Z sm 20 +7 _
tan 20 tt •
_COS 20-1 I -COS 20
=Z — , r . +7«; tan0= — : — -r 2 -;
sin 20 ' ^ sin 20
7^=7,— Z tan 0.
Similarly the moment of inertia about axis B B at right
angles to axis A A is found to be
/fl^iy+Z tan 0.
8o. Oblique Loading. — When the plane of loading on a
beam does not coincide with one of the principal axes of the
section the beam does not deflect in the plane of the loads and
the neutral axis is oblique to that plane. In a cross-section of
such a beam the stress at a point whose coordinates are a and
b referred to the A and B axes may be found by resolving the
tig.*
bending moment at the section into its components in the direction
of each principal axis, finding the stress at the point due to each
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MOMENTS OF INERTIA OF PLANE AREAS. 79
component and adding algebraically. Thus the stress at the
point D, Fig. 40, due to a bending moment, Af , is
(M sin 0)b (M cos 0)a
/- T. + T-
It may sometimes be necessary to find the direction of the
neutral axis to determine which is the extreme or most stressed
fibre. Since there is no stress at the neutral axis, if the last
equation be made equal to zero, a and b will become coordinates
of some point on the neutral axis, and the ratio between them
will be the tangent of the angle, /?, which the neutral axis makes
with the A axis,
b Ia
tantf=- = -T^ctnfl.
r a I*
Example. — A 6X4X$-in. angle is used as a purlin on a roof of
I pitch (tan _1 =$), as shown in Fig. 40. What is the maximum fibre
stress due to a bending moment M? Centre of gravity lies 0.99 in.
from back of longer leg and 1.99 in. from shorter.
/«= 17.40 in. 4 , /„=6.27 in. 4 .
2=(iX4)(2-o.99)(i.99-o.25)-h(iX5.5)(-o.99+o.25)(-3.2S+i.99)
«+ 6.08 in. 4 .
tan 2$= 2X6.08 -^(6.27— 17.40)= — 1.092,
0= — 23 46', tan $= — 0.440.
7^=17.40— 6.o8(— 0.440)= 20.07, /b = 6.27+6.o8(— o.44o)=3.6o.
0=tan -1 1.5— <£=8o°o4 / , sin^o.985, cos0=o.i73, ctn0=o.i75.
tan/?= — 20.07X0.1754-3.60=— 0.978, £=— 44 02'.
Laying off angles and scaling gives
0=1.98 in., £=3.05 in.
/=M(o.985X3.o5 4-20.07)+ 3/(0.173X1.98 4- 3.60) =0.2433/.
If the load is applied along the Y axis,
fl=9o° -0=ii 3 ° 46',
sin 0=0.915, cos 0=— 0.403, ctn0=— 0.440.
tan £=20.07X0.4404-3.60= 2.460, £=67° 53'.
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So STRUCTURAL MECHANICS.
The greatest stress is found on the inside edge of the lower leg, where
a= + 1.18 in., 6= —3.88 in.
/=- Jf(o.9i5X3.88^2o.o7)-Jf (0.403X1. 18-^3.60)= -0.309J/.
Examples. — 1. Find the moment of inertia of a trapezoid, bases a
and b, height h> about one base.
2. A 12-in. joist has two mortises cut through it, each 2 in. square,
and 2 in. from edge of joist to edge of mortise. How much is that
section of the joist weakened ? 2 \ or 2 ^%.
3. A bridge floor is made of plates rolled to half -hexagon troughs,
6 in. face, 5.2 in. deep, 12 in. opening, £ in. thick. Find the resisting
moment of a section 18 in. wide. 20.8/.
4. If that floor is 14 ft. between trusses and carries two rails 5 ft
apart, each loaded with 2,000 lb. per running foot, what will be the
unit stress? 7>790 lb.
5. Six thin rolled shapes, web a, make a hexagonal column, radius
a, with riveted outside flanges, each b in width. Prove that
i2(a+2&)
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CHAPTER V.
TORSION.
81. Torsional Moment. — If a uniform cylindrical bar is
twisted by applying equal and opposite couples or moments at
two points of the axis, the planes of the couples being perpen-
dicular to that axis, the particles on one side of a cross-section
tend to rotate about the axis and past the particles on the other
side of the section, thus developing a shearing stress that varies
with the tendency to displacement of the particles, that is, directly
as the distance of each particle from the centre. The unit shear
then is constant on any r ing, and the shearing stresses thus set
up at any section make up the resisting moment to the torsional
moment of the applied couple. As all cross-sections are equal
and the torsional moment is constant between the two points
first referred to, each longitudinal fibre will take the form of a
helix.
82. Torsional Moment of a Cylinder. — If the unit shear at
the circumference of the outer circle, Fig. 41, of radius R\ and
diameter d is ji, the value at a distance R from the centre will
be, by the above statement, q=q x R+Ri. The total shearing
force on the face of an infinitesimal particle whose lever-arm is
R, and area RdRdO, will be -^-R 2 dRdO, and its moment about
-Ki
the centre will be -jrR 3 dRdO. Hence the resisting moment for
-Kl
a cylinder is
r-?r / * / 2 ^ 3 ^^=^-=- L ^i^i 3 =o.i 9 6g 1 (i 3
K\Jo Jo K\ 2
81
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8a
STRUCTURAL MECHANICS.
Hence the resisting moment against torsion resembles in
form the resisting moment against flexure, but differs in using
the polar moment of inertia of the cross-section for the rectangular
one, and in having qu maximum unit shear, in place of /, maxi-
mum unit tension or compression.
As the rings of metal situated farthest from the centre of a
shaft offer the greatest resistance to torsion, it is economical of
metal to make the shaft hollow. If d is the internal and D the
external diameter of a shaft, the resisting moment is found by
subtracting the moment of a shaft of diameter d from that of
a shaft of diameter D, hence
T^o.ig6qi
D*-d*
83. Torsional Moment of a Square Shaft. — If a square bar
s twisted and the shear is assumed to vary on the cross-section
>-p^
rtea
Fig. 43
with the distance of the particles from the centre, / -$& 4 , r\ -&\/i
and
This assumption is not correct. The unit shear is actually the
greatest at the middle of each side. For rectangular sections the
preceding treatment would be seriously in error, but for a square
section the error is not important. The last Coefficient should
be about 0.208. For shafts the cylindrical form is now almost
universal. See § 85.
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TORSION. 83
The exact investigation for sections other than circles is very
involved. See "Theory of the Elasticity of Solid Bodies,"
Clebsch, translated from the German into French, with Notes
by de St. Venant and Flamant; and report of Chief of Engineers,
U. S. A., for 1895, p. 3041, Part IV.
Example. — Design a round shaft to transmit 500 H.P. at 100
revolutions per minute if £1=8,000 lb. per sq. in.
500X33,000-^100= 165,000 ft.-lb. of work per revolution.
Assume this work to be performed by a force P rotating about the
centre of the shaft at a distance unity. The work done by P in one
revolution will be the force into the distance through which it acts,
or 27rP, and the torque, T, will be PXi; hence
7 = 165,000-?- 27: =26,300 ft.-lb. = 315,000 in.-lb.
j* 315,000 , , . .
o 3 = — —rrz =201, a=6 in. nearly.
0.196X8,000
84. Twist of a Cylindrical Shaft. — If the surface of a cylin-
drical shaft is divided into small squares by two sets of lines,
one set running along the elements and the other at right angles
to them, the first set will become helices when the shaft is twisted,
while the second set will remain unchanged with the result that
the squares will become rhombs, as is shown in Fig. 42. The
angle of distortion of the rhombs is <f>, which is the angle that
every tangent to a helix makes with the axis of the shaft. When
the free end of the shaft has rotated by the fixed end through
an angle 0, for small distortions, such as occur in practice, <f> =—7—.
By definition of the shearing modulus of elasticity, § 173,
r £1 g£ 2qj 16T
n 2q x l 32TI io77
"-cJ"^-c^' nearly '
Example. — A round shaft 2$ in. diameter carries a pulley of 30 in.
diameter; the difference in tension on the two parts of the belt is
2,000 lb. Then T =2, 000 -15=30,000 in.-lb. if the torsional moment
is entirely carried by the section of the shaft on one side of the pulley.
If the shaft is 30 ft. long and C= 11,500,000,
~ 10-30,000- 30- 12 -2 4
0— - 2 — = o. 240.
1 1,500,000 «5 4
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84
STRUCTURAL MECHANICS.
To reduce this angle to degrees multiply by 180-7-71=57.3 and obtain
13 46'. q\ =9,800 lb. per sq. in.
85. St. Venant's Equations for Torsion. — When a torsional
moment is applied to a body whose section is not a circle, the
following equations have been given by M. St. Venant for the
maximum unit stress produced, which is found at points of the
boundary nearest the centre:
3 Tb .
<7i =— -J — for a rectangle whose shorter side is b and whose
moment of inertia is taken through the centre of gravity about
an axis parallel to the longer side.
rrt
ji =— -jb for an ellipse whose least semidiameter is b and
whose moment of inertia is taken about the greatest diameter.
The torsional angle for unity of length
TJ TJ
= about 40:™, for rectangle, and =47: 2 -^, for ellipse;
where 7= polar moment of inertia about axis through the centre
of gravity, and 5= cross-section.
86. Combined Bending and Torsion. — A shaft is often
subjected to bending in addition to torsion, and in such a case
must be designed to resist both. The normal unit stress on
the cross-section at the extreme fibre from the bending moment
on the beam or shaft is /, tension on one edge, compression on
the opposite edge.
The unit shear on the same cross-section at the same extreme
fibre is ji, and as the shears on two planes at right angles are
rO-nV)
Fig. 43
N *
equal by § 154, the stresses at the extreme fibre on a longitudinal
( and on a transverse plane are as shown in Fig. 43. (The plane
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TORSION. 85
of the bending moment is at right angles to the plane of the
paper.) The principal stresses can be found by the method of
§ 168.
AB is the plane of cross-section; -NO =/; RN^ji. Then
RO is the resultant unit stress on AB, R'O = ji on second plane
revolved 90 to make the two planes and normals coincide. Draw
RR' connecting the extremities of the two stresses. As its middle
point falls on the middle point, M, of NO,
/>!=OM + MR, /> 2 = OM-MR. MR 2 =MN 2 + NR 2 .
pi-V+^P+qi 2 , (x)
/>2=i/-^i/ 2 + ?l 2 .
By § 152, qmax-Kpi-h) »^i/ 2 +?i 2 -
Since M=*fl+yi and T = qiJ+yi and since 2/=/ for a
square or a circle, (1) may be multiplied by / -s-yi and transformed
to
Mi-KM + y/Mt + T 2 ), (2)
in which M = original bending moment at the section, T= original
torsional moment, and Mi — equivalent resulting bending moment
for which the shaft should be designed so that the unit stress
shall not exceed / when both M and T occur at the given section.
If (1) is multiplied by J +yu there results
T x =MWM 2 + T 2 (3)
as an equivalent torsional moment in which Ti^piJ +yi.
Although neither equation (2) nor (3) is rational, since pi does not
act on a right section, the conception of an equivalent bending
moment is less objectionable than the conception of an equivalent
torsional moment because pi is a normal stress. But in designing
shafting it is usual to employ the same working value for unit
fibre stress, /, as for unit shear, q x \ consequently the results
obtained are the same whether equation (1), (2), or (3) is used.
If the deformation of the material is taken into account as in
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86 STRUCTURAL MECHANICS.
§175, the new principal stresses become Pi'—pi— \p% and
tf-h-lpx-
Pi'-lf+tV tf + qi*,
Ti'-lM+lVAP + T 2 .
Examples. — 1. If the pulley of the previous example weighs 500 lb.
and is 12 in. from the hanger, on a free end of the shaft, and the un-
balanced belt-p ull of 2,000 lb. is horizontal, the resultant force on the
pulley is 5oov^i 2 +4 2 = 2,060 lb., which causes a bending moment of
24,720 in.-lb. at the hanger. Then
-Wi =s it 2 4>720+\/(24,72o 2 -f30,ooo 2 )]=3i,8ooin.-lb. >
which will cause a fibre stress of
. 3i,8ooX2 3
* 0-196X5* " IO,4 °° lb - *** Sq ' ^
2. The wooden roller of a windlass is 4 ft. between bearings. What
should be its diameter to safely lift 4,000 lb. with a 2-in. rope and
a crank at each end, both cranks being used and / being 800 lb. ?
8* in.
3. Design a shaft to transmit 500 horse-power at 80 revolutions
per min., if q\=* 9,000 lb. d = 6 in.
4. How large a shaft will be required to resist a torsional moment
of 1,600 ft.-lb. if 01 = 7,500 lb.? If the shaft is 75 ft. long and C**
11,200,000, what will be the angle of torsion? 2§ in.; 30$°.
5. What torsional moment will a hollow shaft of 5 in. internal
and 10 in. external diameter transmit if q\ = 8,000 lb. per sq. in.?
What is the size of a solid shaft to transmit the same torsional moment ?
What is the difference in weight between the two shafts if a bar of steel
of one inch section and one foot long weighs 3.4 lb. ?
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CHAPTER VI.
FLEXURE AND DEFLECTION OF SIMPLE BEAMS.
87. Introduction. — As the stresses of tension and compression
which make up the resisting moment at any section of a beam
cause elongation and shortening of the respective longitudinal
elements or layers on either side of the neutral plane, a curvature
of the beam will result. This curvature will be found to depend
upon the material used for the beam, upon the magnitude and
distribution of the load, the span of the beam and manner of
support, and upon the dimensions and form of cross-section.
It is at times desirable to ascertain the amount of deflection, or
perpendicular displacement from its original position, of any
point, or of the most displaced point, of any given beam carrying
a given load.
Further, the investigation of the forces and moments which act
on beams supported in any other than the ways already discussed
requires the use of equations that take
account of the bending of the beams J\
under these moments. There are too
many unknown quantities to admit of a
solution by the principles of statics alone.
The required equations involve expres-
sions for the inclination or slope of the
tangent to the curved neutral axis of
the bent beam at any point, and its
deflection, or perpendicular displacement,
at any point from its original straight
line, or from a given axis. The curve
assumed by the neutral axis of the bent beam is called the elastic
curve.
87
/
IP
Fig. 44
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8& STRUCTURAL MECHANICS.
88. Formula for Curvature. — If, through the points A and
B, on the elastic curve, Fig. 44, and distant ds apart, normals
C D and K G to the curve of this neutral axis are drawn, the
distance from AB to their intersection will be the radius of
curvature p for that portion of the curve. If through A a plane
F H is passed parallel to K G, the distance F C will be the elon-
gation, or H D will be the shortening, from the unit stress /, of
the extreme fibre which was ds long before flexure. Cross-
sections plane before flexure are plane after flexure, § 62.
AO=jo; AC«=yi; CF = ^ds, § 10. From similar triangles
ACF and O A B, pids^yxi^ds, or /> = -p As, by §64,
M — — t — =^7> the reciprocal of the radius of curvature, called
y x p EI r
the curvature or the amount of bending at any one point.
89. Slope and Deflection. — If the elastic curve is referred to
rectangular coordinates, x being measured parallel to the original
straight axis of the beam, and v perpendicular to the same, the
calculus gives for the radius of curvature
[■♦®7
** d*v
dx 2
For very slight curvature, such as exists in practical, safe beams,
-7- is a very small quality, and in comparison with unity its square
may be neglected. Then
p~dx*~EV
As M is a function of x, as has been seen already, the first
dv
definite integral, -7-, will give the tangent of the inclination or the
slope of the tangent to the curve of the neutral axis at any point
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 89
x, and the second integral will give v, the deflection, or perpen-
dicular ordinate to the curve from the axis of x. Thus
dv
^=slope, *;
di cPv
El -j- = moment, M=EI-t—£
dM , „ rf 3 !/
-^=shear, ^=£7^;
g-load, V-EI&.
While the following applications of the operations indicated
in the last paragraph are examples only, the results in most of
them will be serviceable for reference.
The student must be careful, in solving problems of this class,
to use a general value for M, and not M maximum. The origin
of coordinates will be taken at a point of support, such a point
being definitely located ; x is measured horizontally, v vertically,
and —v denotes deflection downwards.
The greatest deflection for a given load is v max . The greatest
allowable deflection for a fibre stress / is v x .
If M+I is constant, the beam bends to the arc of a circle.
This happens where M is constant and J is constant, or where
J varies as M.
Example. — The middle segment of a timber beam carrying a load,
a FI
W, at each quarter-point bends to the arc of a circle. M = \Wl\ p= -rrzj.
If the beam is 6 in.Xi2 in.X2o ft., JF= 2,000 lb. and £=1,500,000
lb. per sq. in., then 7= T V-6-i2 3 = 864 in. 4 and p=- — — — !
r ^ » n r 2,000-2012
= 10,800 in. =900 ft.
90. Beam Fixed at One End; Single Load at the other; origin
at the wall; length =/. Let the abscissa of any point on the
elastic curve be x, and the ordinate v. At that point
M x =-W(l-x).
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9 o
Then
STRUCTURAL MECHANICS.
The slope at the point x is found by integrating.
dv W
At the point where x=o, the slope is zero and therefore C=o.
Integrating again,
W
When x=o t v=o, therefore C'=o. The equation just found is
the equation of the elastic curve. If x is made equal to / in the
above equations, the slope and the deflection at the end of the
beam are found to be
• WP
tmax - ~ 2 EV
WP
*^mnx..
To determine the maximum allowable deflection of a given
beam consistent with a safe unit stress in the extreme fibre at the
section of maximum bending moment, substitute, in the expression
for v maXt9 the value of W in terms of /.
M max, = -
WlJl
WP }P
• wr — 1,
V i = -T^7 =
3EI 2> R y\
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS.
Example. — If /= 60 in., 6=4 in., h=8 in., PF=8oo lb., £=
r 4'8 3 512 . 800-602-3
I = = - — ; max. slope = — = 0.006;
2- 1. 400.000 • CI 2
91
2 1,400,000 51 2
I,400,000;
Umax. =
— = 0.24 in.; and if /= 1,200 lb., maximum safe de-
3 -1,400,000- 512
a ^. i^ooXoo 2 . .
flection= —— — — °- 2 ° m.
3X1,400,000X4
It will be seen that, for a given weight, the maximum bending
moment varies as the length /; the maximum slope varies as Z 2 ,
and the maximum deflection as P. The slope and deflection
also vary inversely as J, or inversely as the breadth and the cube
of the depth of the beam. The maximum safe deflection, how-
ever, consistent with the working unit stress /, varies as Z 2 , and
inversely as yi, or the depth of the beam. These relationships
are true for other cases, as will be seen in what follows.
The ease with which problems regarding deflection are solved
depends greatly upon the point taken for the origin, as it influences
the value of the constants of integration.
91. Beam Fixed at One End; Uniform Load of w per unit
over the whole length, /; origin at the wall.
M x =-\w{l-x) 2 .
dx
C=o.
w
When #=0, *=o;
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9 2 STRUCTURAL MECHANICS.
When #=o, v=o; .'. C'=o. When #=/,
(wl)P
Vina*-
SEI
For maximum safe deflection consistent with unit stress, /,
in the extreme fibre at the dangerous section,
(wl)P IP
8£/ 4£yi'
92. Combination of Uniform Load and Single Load at one
end of a beam fixed at the other end. Add the corresponding
values of the two cases preceding.
M max = -[wi+i(wQil;
Vmax. = ~
Note, in the expression for i'^,, the relative deflections
due to a load at the end and to the same load distributed along
the beam; and compare with the respective maximum bending
moments.
Example. — If the preceding beam weighs 50 lb., the additional
deflection will be =0.005 m -> to ° small a quantity to
8-1,400,000-512 -1 j
be of importance. In the majority of cases the weight of the beam
itself may be neglected, unless the span is long.
93. Beam Supported at Both Ends; Uniform Load of w
per unit over the whole length /; origin at left point of support.
M z = \w(l— x)x.
d?v w _
dx 2 ~ 2 EP lX X);
dv w
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS
93
From conditions of symmetry it is evident that the slope is
zero when x = J/; hence, to determine the constant of integration,
make
As v»o when #=o or /, C'=o. When #=J/,
5 (w/)/ 3
* w ~ 384 EI #
For maximum safe deflection consistent with a unit stress,
/, in the extreme fibre at the middle section,
v yi yi/ 48£yi
Examples. — A pine floor-joist, uniformly loaded, section 2X12 in.,
span 14 ft. = 168 in., has deflected $ in. at the middle. Is it safe?
£=1,500,000. By the last formula,
4 2. T9 2
0/ r'» /= 2,300 lb.,
4 48 '1,500,000- 6' ' ,0 '
and the beam is overloaded.
What weight is it carrying? By formula for v max%%
>i3 . T o3
3 ^ . I4*I2°I2 .
-= -^u'/ *; w/= 5,248 lb.
4 384 1,500,000- 2 -I2 3 ' ^ *
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94 STRUCTURAL MECHANICS.
94. Beam Supported at Both Ends ; Single Load W at middle
of span /; origin at left point of support.
d*v W
dx*~2EI X '
This expression will apply only from #=0 to x=\l\ but,
as the two halves of the deflection curve are symmetrical, the
discussion of the left half will suffice.
dv W
When *=J/, t-o. .". C= -\P.
W
When #«o, v-o. .\ C'=o. The limit x=*l is not applicable.
Vma *= 4 8Ef
For maximum safe deflection consistent with a unit fibre
stress,/, JW7-?; W=^r; v 1 ^- fP
yi yj' i2Eyi
Example. — A 10-in. steel I beam of 33 lb. per ft. and /= 162; span,
15 ft., = 180 in., carries in addition a uniform load of 767 lb. per ft.
of span and 6,000 lb. concentrated at the middle. What will be its
deflection and the maximum unit fibre stress?
/5-8oo-i5 , 6,ooo\ 180 3
W ° l-m- + -l8-; 29 ,ooo,ooo-i6 3 °°- 3S m - ;
/^ = (8o^5 + 6^o) l8o . /=l6)66?lb .
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS.
95
95. Single Weight on Beam of Span /, supported at both
ends; IF at a given distance a from the origin, which is at the
left point of support.
Fig. «
As the load is eccentric, the curve of the beam is unsym-
metrical, and equations must be written for each portion, x<a
and x>a.
ON LEFT OF WEIGHT,
M x
X,
W
(Pv
W
v ~EU (ilx '~ iox'+Cx+C").
(1)
v=o, when x— o; .*. C"—o.
ON RIGHT OF WEIGHT.
„ Wa „ Wa„ N
d?v W , . x
_ = _ (fl/ _^).
g.^oto-ja^+CO-rf. (2)
IT
v— o, when #-*/;
For equations to determine the constants C and C, use the
value #=a, when it will be evident that i for the left segment
must give the same value as does i for the right segment, and
v at a must be the same when obtained from the left column as
when obtained from the right. Therefore, from (1) and (2),
v at a gives
C'=-\a*-\aP.
dv W_/t^a 3 .a?l__£_oP\
dx~EIl\ 2*2 6 3/'
c afe aPx \
'" 6" 3/"
dv
dx
W I , ax 1 a 9 aP\
-BR^-T-t-j)-
_ W /alx* ax* a*x aPx a*l\
V ~EIl\ 3 6.6 36/
-^fl-x^-( 3 l-x)x).
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96 STRUCTURAL MECHANICS.
As a is assumed to be less than \l, and the substitution of
dv
x=a in the value of -r- on the left gives a slope which is negative,
the point of v ntax , will be found on the right of W, and for that
dv
value of x which makes -7- on the right zero. Hence
\ax 2 — alx + ia 3 + \aP = o.
*-/-\/i(/»-a*),
which is the distance from the left-hand support to the point of
maximum deflection. Substitute in the expression for v on the
right, to obtain the maximum deflection.
It should be noticed that, when the weight is eccentric, the
point of maximum deflection is found between the weight and
the- mid-span, and not at the point of maximum bending moment,
which latter is under the weight.
96. Two Equal Weights on Beam of Span /, supported at
the ends; each W, symmetrically placed, distant a from one end.
This case may be solved by itself, but can be more readily
treated by reference to § 95. Thus the maximum deflection
will be at the middle, and can be found by making x=\l in the
above value of v for the right segment and doubling the result.
Then
Wa I OJOX Wa m
The deflection under a weight will be given by the addition
of v at a and v at (I -a) of the preceding case. Thus
Wa*,, .. Wa* /n „
Wa 2
vatW=-£-gj(3/-4<z).
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 97
Example. — A round iron bar, 12 ft. long and 2 in. diameter, carries
two weights of 200 lb. each at points 3 ft. distant from either of the
two supported ends. The deflection at a weight =
-( 3 .i2 2 -4-3-i2)=o.56 in.
6 -28,000,000 -22
200 • "? • 1 2 * 7 • 4
The maximum unit bending stress is — = 0,160 lb.
o 22
DEFLECTION OF BEAMS OF UNIFORM STRENGTH.
It will be apparent that a beam of uniform strength will not
be so stiff as a corresponding beam of uniform section sufficient
to carry safely the maximum bending moment; for the stiffness
arising from the additional material in the second case is lost.
97. Uniform Strength and Uniform Depth. (See § 71.) —
Since M=njbh 2 and varies as b h 2 , and 7=n'iA 3 and varies as
bh 3 , M -t-I varies as 1 -*- h. But if h is constant, M +1 is constant
cPv
and j~2 is constant. Therefore all beams of this class bend to
the arc of a circle.
I. Beam fixed at one end only, and loaded with W at the other.
Fig. 22.
If I is the moment of inertia at the largest section, which is
in this case at the wall, and / is the variable moment of inertia,
dx*~Er EP l X) ~ EIo
dv Wl „
te = ~Er x+C; c==a
Wl
v~—nr&+Ci C'=o.
2EI0
WP
Vmax. - 2EIq >
a deflection 50 per cent, in excess of that of the corresponding
uniform beam, while the maximum slope is twice as great.
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9 8
STRUCTURAL MECHANICS,
Examples. — If a triangular sheet of metal, like the dotted triangle
in Fig. 49, is cut into strips, as represented by the dotted lines, and
these strips are superimposed as shown above, the strips, if fixed
at the ends, and subjected to W as shown, will tend to bend in arcs
of circles, and will remain approximately in contact. If /=io in.,
6=4 in., h= J in., W=4oo lb. and £=28,000,000, the deflection will be
400-io 3 -4 3 -i2
2' 28,000,000 '4
An elliptical steel spring 2 ft. long, of 4 layers as shown, each
2 in. broad and £ in. thick, under a load ot 100 lb. at its middle, will,
.* ~ 1 a ^ IOO-I2 3 -8 3 12 . _
if £=20,000,000, deflect 0=2.3 in. The maximum unit
y 2-29,000,000-8 °
50-12-6-
fibre stress will be
8
-=28,800 lb. Note that one-half of the
weight is found at each hinge, and that the deflection of one arm is
doubled by the use of two springs as shown.
II. Beam fixed at one end only and uniformly loaded with
w per unit. Fig. 25.
I _b _ (l-x)* (l-x)* P
Io~b ~ P '
<Pv
w
I
wP
/o
dx* 2 £/ (/ * )2 ~ 2£Jo'
dv
dx~
wP
2EIo
wP
x+C;
4EI0
wl*
4EI0
x*+C;
C=o.
C'=o.
a deflection twice that of the corresponding uniform beam.
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 99
In these two cases there are no constants of integration! since
-J- and v=o when #=o.
dx
m. Beam supported at both ends and carrying W at middle.
Fig. 27.
I:I =b:b =x:%l.
<Pv Wx Wl
dx 2 ~2EI~4EI ;
dv Wl , ^
Slope -o when #=J/; .'. C= -£/.
*-^(i*-V*)+C'l C'-o.
32^/0
a deflection 50 per cent, greater than for a corresponding uniform
beam.
IV. Beam supported at both ends, and uniformly loaded
with w per unit of length. Fig. 29.
I:Io = b:b =x(l-x):iP.
<Pv w „ ON wP
dx 2 2£/ V ' 8EI '
v=^(i^-^+C0; C'=o.
64^/0
a deflection 20 per cent, greater than for a corresponding
uniform beam.
98. Uniform Strength and Uniform Breadth. — In these
cases, as b is constant, I:Io=h 3 :ho z or /=/ r^«
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IOO STRUCTURAL MECHANICS.
V. Beam fixed at one end only, and loaded with W at the
other. Fig. 24.
Ao 2 ~ I ' V = ° /» •
^2= -£7('-*)= -£/;('-*) *;
2 u>7l /**
or twice the deflection of a corresponding uniform beam.
VI. Beam fixed at one end only and uniformly loaded with
w per unit of length. Fig. 26.
^ /-* T T (l-x)*
h ~ I ' A ~ l0 P '
^=- 2 £7 (/ - :v)2= -2£7o (/ - ar) " I;
*r 2 w Iog(/ ~* ) - Iog/1 -
w/ 3 /■'
Vfnax= ^ET J o Dog(/-*)-log(]cte*
if/ 3 r n'
= IeTL* Iog (^ -*) -* _/ lo 8 (' -*) -* lo s m
if/ 3 w/ 4
= tfZfV log o-/-/ log o-/ log /+ / log /) = -^gj",
or four times the deflection of a corresponding uniform beam.
* Log (/— #)=- «; <te=(/i/; #=v; du=> — r—.
/. yiog (/-*)<**=* log (/-*)+ /jz-jlx- By division, ~^=-i+—;
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS. IOI
VII. Beam supported at both ends and carrying W at middle.
Fig. 28.
A 2 x ,.18*3
7-5=77. i=i \J-p-.
ho 2 V
d*v Wx Wt*
:-*
dx 2 2EI 4VIEI0
dv Wfl , . „,
dx=WlETo (21 * +C) '
Whenx-J/, t-=o, and C- -2V / J/.
v^=-%£ r f\xi-Vy)dx
2V2EI J o
Wl* 1 2 ft l* \ _ WP
,B 2\/^o\3 2V2 2V2/ 24EI0
or twice the deflection of a corresponding uniform beam.
VIII. Beam supported at both ends, and uniformly loaded
with w per unit of length. Fig. 30.
h 2 x(l-x) SV^/-*) 3
ho 2 " \P ' ° P
<Pv w , ox wP „ ox .
_ = _ ( ^_^) = ___ (to _^-» ;
Jv a// 3
(versin~ 1 -i- + Cj;
dx 16EI0
dv
-t-=o, when # = £/; .*. C= — versin" 1 1= — Jtt.
w/ 3
i6£/ ,
wP
I fversin" 1 -^-— \n\dx
r 2# "1 ^'
i6£7 L^~^ versin" 1 y+\/(/A;-^ 2 )-i^J
w/ 3 ^/ n "I _ 0.2854W/ 4 _ o.oi8u>/ 4
"LI 4 J ^
16JS/0L2 4 J i6£/ £/o
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1 02 STRUCTURAL MECHANICS.
or 37 per cent, greater deflection than for a corresponding uniform
beam.
Other beams might be analyzed where both b and h varied
at the same time. The method of analysis would agree with
the above; but the cases are not of sufficient practical value to
warrant their discussion here.
99. Sandwich Beam. — If a beam is made up from two
materials placed side by side, as when a plate of steel is bolted
securely between two sticks of timber, the distribution of the load
between the several pieces can be found from the consideration
that they are compelled to deflect equally. If the pieces are
of the same depth, the extreme fibres of the two materials must
undergo the same change of length; or by § 10, if the subscripts
w and s stand for wood and steel respectively,
s * Jw I a Jw &W
The resisting moment of such a beam is
Jf-|/.»wW+i/ f 6 i W-i/ li (^+^)w.
Such beams are easily figured by substituting for the steel an
equivalent breadth of wood and proceeding as if the beam were
entirely of wood or vice versa.
If E for timber is 1,400,000 and for steel 28,000,000, the ratio
of the stresses will be *V; and if } w =800 lb., /• = 16,000 lb. on the
square inch.
Example. — Two 4Xio-in. sticks of timber with a ioXj-in. steel
plate firmly bolted between them will have a value of
,_ (800 • 8 + 16,000 -i)io? . 1U
If-* ^ ^ — = i73>333 m.-lb. f
the plate supplying -^ of the amount. The combination for a span
of 10 ft. would safely cany =5,778 lb. load at center, or 11,555
lb. distributed load, in place of 3,555 or 7,110 lb. for the timber alone.
"\
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 103
zoo. Resilience of a Beam. — If a beam carries a single weight
W, and the deflection under that weight is v it the external work
done by that static load on the beam is iWvi. If this value of
v\ is that which causes the maximum safe unit stress /, the quantity
\Wv\ is known as the resilience of the b am, or the energy of the
greatest shock which the beam can bear without injury, being
the product of a weight into the height from which it must fall
to produce the shock in question. For a beam supported at
both ends, loaded in the middle", and of rectangular section,
* 1 = 6£A' and W = -jT'
x _ 1 2 jbh* fP 1 P ...
2 1 23 / 6Eh 18 E
The allowable shock, or the resilience, is therefore propor-
tioned to P+E, which is known as the modulus of resilience of
the material, and to the volume of the beam. These relation-
ships hold for other sections, and beams loaded and supported
differently. The above formula should not be rigorously applied
to a drop test, unless / is below the yield-point.
Example. — A 2X2 in.-bar of steel, 5 ft. between supports, if /=
16,000 lb., ought not to be subjected, from a central weight, to more
. I 16,00a 2 2 2 ' 60 . . „ .
than — = 118 m.-lb. of energy.
18 29,000,000
If the load is distributed over a similar beam, the deflection
at each point will be v, and the total work done will be \\vwdx.
If w is uniform, and the beam is supported at its ends,
, P l , w 2 P/lx? x* Px\j
wP }I 5 }P
As T =- and ^1=^^
the above expression becomes
w 2 l* 8
Resilience = ^7 = — (wl)Vi.
240EI 25 V
24o£T
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104 STRUCTURAL MECHANICS.
Example. — A weighted wheel of 1,000 lb. drops £ in. by reason of
a pebble in its path at the middle of a beam 3X12 in., 15 ft. span.
If £= 1,400,000 to find /:
External work=i,ooo(H*)=' Li l : ^/ 2 — / 2 .
w ' 18 E 70,000'
= j_ JP_ = i8ojjr8o /= _9_ ;
12 Eyi 6-12-E 28,000
soo+ ^ ,= ^ /2 - /= 2 ' IS ° lb - p" ■* b -
Resulting deflection =0.69 in. Static unit stress would be 625 lb.
and v=o.2 in. In an actual bridge the shock is distributed more or
less in the floor and adjacent beams.
10 1. Internal Work. — The internal work done in a beam may
be divided into two parts, that of extending and compressing the
fibres, and that of distorting them, the, first being due to the
bending stresses, and the second to the shearing stresses. The
work of bending will be found first.
Let the cross-section be constant. The unit stress at any
point of a cross-section = ^; the force on a layer zdy = pzdy. The
elongation or shortening of a fibre unity of section and dx long
by the unit stress p=pdx+E. The work done in stretching or
1 f- f M
shortening the volume zdydx=---g-zdydx. But />= — y=-jy*
The work done on so much of the beam as is included between
two cross-sections dx apart will be
1 M 2 /*+» M 2
7Ei 2dx J^ ^y-iEf 1 *-
Substitute the value of M for a particular case in terms of
x and integrate for the whole length of the beam. Thus for a
beam supported at ends and loaded with W at the middle, M = \Wx
at any point distant x from one end for values of x between o
and £/. Then
2 r* W 2 P l
IP/ 3
4EIJ ~"~~tf>Er
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FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 105
The internal work due to shear is not readily found unless
a simple form of cross-section is chosen. If the section is rect-
angular the shear at any point on the cross-section is, by § 72,
The distortion in a length dx is qdx+C and the work done upon
q qdx
sl volume zdydx is 77- -zdy. The work done upon the volumes
2 o
between two sections dx apart is
dx /*+** 18F 2 /*+** 1 F 2
For a beam carrying a single load at the centre, as above, -F= \W f
a constant, and the work done by the shearing forces is
3 WH
20 b hC
102. Deflection Due to Shear. — In the cases of flexure so far
treated the bending moment only has been taken into account
in finding the deflection. The shearing stresses, however,
cause an additional deflection which is generally too small to be
of practical account. The deflection of a rectangular beam
carrying a single load at the centre can be found from the results
of the preceding section since the internal work must be equal
to the external
iWv ~ 9 6EI + 2obhC'
_ WP 3WI
V ~4&El' h iobhC m
The first of these terms is the deflection when the shear is not
taken into account, and it has the same value as was obtained
in § 94. The ratio of the second term to the first is r „ t which
shows that in rectangular beams of ordinary proportions the
shearing deflection is but a small proportion of the whole.
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106 STRUCTURAL MECHANICS.
Examples. — i. What is the deflection at the middle of a aXia-in.
pine joist of 12 ft. = 144 in. span, supported at ends and uniformly
loaded with 3,200 lb.? £=1,600,000. 0.27 in.
2. What is the deflection if the load is at the middle? 0.432 in.
3. Find the stiff est rectangular cross-section, bh, to be obtained
from a round log of diameter d. b=bd.
4. A 4X6-in. joist laid flatwise on supports 10 ft. apart is loaded
with 1,000 lb. at the middle. The deflection is found to be 0.7 in.
What is E? 1,607,000.
5. What is the maximum safe deflection of a 12-in. floor- joist, 14 ft.
span, if /= 1,200 lb. and JE= 1,600,000?
0.37 in. for uniform load; 0.29 in. for load at middle.
6. What is the diameter of the smallest circle into which J-in.
steel wire can be coiled without exceeding a fibre stress of 20,000 lb.
per sq. in. ?
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CHAPTER VII.
RESTRAINED AND CONTINUOUS BEAMS.
103. Restrained Beams. — When a beam is kept from rotating
at one or both points of support, by being built into a wall, or
by the application of a moment of such a magnitude that the
tangent to the curve of the neutral plane at the point of
support is forced to remain in its original direction (commonly
horizontal) at such point, the beam is termed fixed at one or
both supports. The magnitude of the moment at the point of
support depends upon the span, the load and its position. It
is the existence of this at present unknown moment which calls
for the application of deflection equations to the solution of such
problems as those which follow, there being too many unknown
quantities to permit the treatment of Chapter III.
In applying the results obtained in the following cases to actual
problems, one should feel sure that the beam is definitely fixed
in direction at the given point. Otherwise the values of M y F>
and v will be only approximately true.
104. Beam of Span /, Carrying a Single Weight W in the
middle and supported and fixed at both ends. Origin at left
support. Fig. 50.
The reactions and end moments are now unknown. The
beam may be considered either as built in at its ends (as at the
right in above figure), or as having an unknown couple or moment
Qb applied at each point of support (as at the left), of a magnitude
just sufficient to keep the tangent there horizontal.
The reaction at either end will then be \W + Q, while the
shear between the points of support will still be±$W. For
values of x<\l,
M x =-Q(b+x) + (iW+Q)x=$Wx-Qb.
107
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io8
STRUCTURAL MECHANICS.
If this value is compared with that of M x in § 94, it is seen
that a constant subtractive or negative moment is now felt over
the whole span in combination with the usual %Wx.
Slope is zero when #=0; .'. C=o. Also slope is zero when
o=^WP-lQbl. -()& = -JW7,
the negative bending moment at either end. That it is negative
appeals by making x=o in M x above.
If this value of Qb is substituted in the first equation, giving
M x = iW(x— £/), the point of contraflexure is located at #=|/,
and the bending moment at middle, where x= J/, is M^^
= -f \Wl } or one-half the amount in § 94. Substituting the value
of Qb in the equation for slope,
w
*=^7(J*3-i/* 2 +c').
As v = o when x = o, C = o. When x = J/,
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RESTRAINED AND CONTINUOUS BEAMS. 109
Vm "-~4EI\24 16/ 1 9 2EI'
The beam is therefore four times as stiff as when only sup-
ported at ends*
. // Wl,-. 8// , }P
**-?"-& and Vl= ^Wi'
so that only one-half the deflection is allowable that is permitted
in § 94, but the beam may safely carry twice the load.
It is useful to notice that this beam has a bending moment
at the middle equal to that which would exist there if the beam
were cut at the points of contraflexure and simply supported
at those points, and that the two end segments, of length £/,
act like two cantilevers each carrying \W, the shear at the point
of contraflexure.
If the weight were not at the middle the moments at the
two ends would differ, equations would be needed for each of
the two segments, and the solution, while possible, would be
much more complicated.
Example. — A wooden beam, 6 in. square and 7$ ft. span, is built
into the wall at both ends. A central weight of 3,000 lb. will give a max.
fibre stress of — —^ — =937i lb. per sq. in. at the middle and both
•2 000 " 00^ * 1 2
ends. The deflection will be — — — 77=0.07 in. if £= 1,500,000.
192- 1,500,000- 6 4 ' J
The allowable deflection for /== 1,200 is — ! — - — = 0.00 in., and
24- 1,500,000-3
max. allowable PF= 3,000 -$=3,860 lb.
105. Beam of Span /, Uniform Load of w per unit over the
whole span, fixed at both ends. Origin at left support. Fig. 51.
As in the previous case, the reaction at either end may be
represented by \wl+Q. The shear at x is \wl— wx, which ex-
pression changes sign at the middle and at either point of support;
hence at those places will be found M^*,
M x =($wl+Q)x-$wx 2 -Q(b+x) = $wlx-~lwx 2 --Qb.
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HO STRUCTURAL MECHANICS.
Compare with § 93.
d?v 1
dv 1
t-=o when #=0; /. C=o. 7~ = ° when #— /;
(fo
wP wP ^ t , ^, w/2
T -Q«=o. -06= — -,
Fig. 51
the negative moment at each point of support. If x= $/,
Af at middle^^i-i-^)^— .
24
Substitute the value of Qb and get
S— 5e^^-^ + ^'
2£/
2hI\I2 6 12 /
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RESTRAINED AND CONTINUOUS BEAMS. Ill
Since v ■= o when x = o, O « o. When # = J/,
which is one-fifth the value of § 93.
The points of contraflexure occur where M x ~o;
The second term is the distance from the middle, each way,
to the points of contraflexure. If M is calculated for the middle
point of a span l+Vfr it will prove to be wP+24, as above.
Since
(wl)l }I . 12 }I
12 y x yil
and V\ =
IP
2>2Eyi
or 0.3 as much as in § 93. The beam may safely carry, however,
50 per cent, more load.
Example. — An 8-in. I beam of 12 ft. span, carrying 1,000 lb. per foot,
if firmly fixed at both ends and not to have a larger unit stress than
12,000 lb., should have a value of /==
I,000'I2'I2'I2'4
= 48. An
12*12,000
8-in. steel beam, 18 lb. to the foot, 7= 57.8, will satisfy the requirement,
the load then being 1,018 lb. per foot. The deflection will be 0.6 in.
106. Beam of Span /, fixed or horizontal at P 2 , supported
at P\ and carrying a uniform load of w per unit. Fig. 52. Origin
at P\ and reaction unknown.
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H2 STRUCTURAL MECHANICS.
It will be seen from the sketch that a beam of length 2/, rest-
ing upon three equidistant supports in the same straight line, will
come under this case.
cPv 1
^= Tl {PiX-\w&);
dv
-i-=o when*=/; /. C=\wP-\P x P.
I [P X X* wx* wPx P x Px \
EI\ 6 24 6 2 /
, v=o when #=0; .'. C=o. If#=/, v=o, and
Substitute this value of P\ in the above equations.
dv
For v max% make j- = o, or f/* 2 — J* 3 — ^P « o.
As a minimum value of v, or v=o, occurs for #=/, divide
by x — I and obtain
&V 2 — Ix— / 2 =o, or #=/ ^-=0.4215/.
Then v^ = -0.0054-^j- .
To find points of M^^ put F x =o, or #=§/.
Also, by inspection, M,^*. when x=l.
For * = f/, M^*. = (■& - T fa)wP = -rf^^/ 2 .
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RESTRAINED AND CONTINUOUS BEAMS. 113
For x=l, M^-d-^wP- -JwP.
For the point of contraflexure §/#— Jx^o, or x=*\l, as
was to be expected from the position of the point of maximum
positive M .
Note again that the point of maximum bending moment
is not the point of maximum deflection.
It will be seen that a continuous beam of two equal spans /,
uniformly loaded with w per unit, has end reactions of § wl and
a central reaction of 2X$wl=$wl; that points of contraflexure
divide each span at \l from the middle pier, and that the bending
moment at the middle of the remaining segment of \l is, as above,
}•! '$wP=ji*wP. It will also be seen that, since the bending
moment at P2 is— fat 2 , a. uniform beam, continuous over two
equal spans, each /, is no stronger than the same beam of span
/ with the same uniform load. It is, however, about two and a
half times as stiff.
Example. — A girder spanning two equal openings of 15 ft. and
carrying a 16-in. brick wall 10 ft. high of no lb. per cubic ft. will
1 1 t 5-4'iiO'iO'i5 .. ,
throw a load of - -= 27,500 lb. on the middle post and
1 1. , 4«iio»io«i5'i5 P ..
must resist a bending moment of — - — - =41,250 ft.-lb.
107. Two-span Beam, with Middle Support Lowered. —
A uniform beam, uniformly loaded and supported at its ends,
will have a certain deflection at the middle which can be calculated.
If the middle point is then lifted by a jack until returned to the
straight line through the two end supports, the pressure on the
jack, by § 106, will be five-eighths of the load on the beam. Since
deflection is proportional to the weight, other things being equal,
if the jack is then lowered one-fifth of the first deflection referred
to, the pressure on the jack will be reduced one-fifth, or to one-half
of the load on the beam. Hence, if a uniformly loaded beam of
two equal, continuous spans has its middle support lower than
those at its ends by one-fifth of the above deflection, the middle
reaction will be one-half the whole weight, the bending moment
will be zero at the middle, and the beam may be cut at that point
without disturbance of the forces.
108. Beam of Span /, fixed at left and supported at right end,
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H4
STRUCTURAL MECHANICS.
and carrying a single weight W at a distance a from the fixed end,
F*g- S3- Origin at fixed end.
Fig. 58
3
The reaction at the supported end, being at present unknown,
will be denoted by P2> and moments will be taken on the right of
any section x. From lack of symmetry, separate expressions
must be written for segments on either side of W.
BETWEEN W AND FIXED END.
M x =P a (l-x)-W(a-x)
dv I
di~*w p ' {lx ~ W) ~ w(ax ~ i * ,)+cl
+Cx+C'~\
— =o when #— o; .\ C— o.
BETWEEN W AND SUPPORTED END.
M x -P % (l-x)
-AW?-?)
+C"*+C"
If #— a, -7- on left" T: on ngnt.
If #~a, v on left— v on right.
v— o when *— o; /. C— o.
.\ C"=TF(a 2 -ia 2 )=-JWa 2 ;
C" = W( - Ja 3 + Ja 3 + Ja 3 ) = $ Wa 3 .
If *=/,vatP 2 =o; .-. P 2 (J/ 3 -J. 3 )-iWo 2 /+i^a«=o,
or
Wa 2
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RESTRAINED AND CONTINUOUS BEAMS. 1 15
If this value of P 2 is substituted in the above equations the desired
expressions are obtained. Thus
Mmax,, by inspection, when # =*o, or x=a.
Wa 2 W
Afma*. — ^-(3*-<0 -Wa- —^a(l-a)(2l-a) at fixed end,
Wa?
and = -^(l-a)(3l-a) at the weight.
The point of contraflexure occurs between W and the fixed
end, where M =o, or
-™(3'-0)(J-*)-(0-*)=o. /. x=al-
2P KO v 2/(/ + a)-a 2
The maximum deflection will be found where t-=o, on the
right or left, according to the value of a.
The above beam may be regarded in the light of two equal
continuous spans with W on each, distant a each side of the
middle point of support.
In solving the more intricate problems in the flexure of beams,
as well as those just treated, each equation of condition can be
used but once in the same problem, and as many unknown
quantities can be determined as there are independent equations
of condition. The reactions and moments at the points of support
are usually unknown, and must be found by the aid of such
flexure equations as have just been used.
Example. — A bridge stringer which is continuous over two
successive openings of 12 ft. each and carries a weight from the
wheels of a wagon of 3,000 lb. at each side of and 3 ft. from the
middle support will be horizontal over that support. Then —M^^
= 2,320.3 ft.-lb. p 2=^7^j-3 2 ' 33 =258 lb. Reaction at middle sup-
port from both spans =2 (3 ,000— 258) =5,484 lb.
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n6
STRUCTURAL MECHANICS.
109. Clapeyron's Formula, or the Three-moment Theorem
for Continuous Loading. — To find the reactions, shears, and
bending moments for a horizontal uniform continuous beam
loaded with W\, W2, W3, etc., loads per running unit over the
successive spans l\> h, h, etc. Fig. 54. P , Pi, P 2 , etc., denote
Fig.M
the unknown reactions; M , Mu M2, etc., the unknown bend-
ing moments at points of support, o, 1, 2, etc.; F , F ly F 2 , etc.,
denote the shears immediately to the right of o, 1, 2, etc.; while
F\> <Py> etc., denote the shears immediately to the left of the
points of support 1, 2, etc.
The origin of coordinates is first taken at O and the sup-
ports are on a level. +M makes the beam concave on the upper
side. As positive shear acts upward at the left of any cross-
section, w is negative.
Consider the condition of equilibrium of the first span 0-1,
or /1, loaded throughout with W\ per unit of length.
Take moments on the left side of and about a section S,
distant x from the origin O. The bending moment at S is
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RESTRAINED AND CONTINUOUS BEAMS. II?
(Pv
M=EI^=Mo+Fox-±wix i . • . . . (i)
Let io> Ht *2'«**n = tangent of inclination of the neutral
axis at o, i, 2, . . . N. Integrate (1) and transpose the con-
stant of integration, io> to the left-hand member and thus obtain
an expression for the difference in slope or inclination of the
two tangents to the bent beam at O and S.
El(J^-i ( ^=M ( p+iF ( p 2 -lw l ofl. *". . . (2)
dv
When x=h, 3~=^i> and hence
Elih-^^M^^F^^w^. • • • (3)
Integrate (2) and determine constant as zero, because v»o
when #=0.
Eliv-i^^lM&z + lFtffi-JfWiX*. ... (4)
Make x=l u then v=vi=o,
and -£/*y , = W<ti 2 + W1 8 -jWi*,
or -£/i =Wi+iW-*Wi 3 (5)
Eliminate i by subtracting (5) from (3).
£Z»i-JJfo/i+iWi 2 -M/i a (6)
If the origin is taken at 1 instead of o an equation like (5)
is obtained for the second span l 2 , or
-E/«i-Jlfi/ a +i-PiV-AwiV (7)
Add (6) and (7), obtaining
o = $Mo/i +iMi/ 2 +i J F /i 2 + ii 5, i/2 2 -i^/i 3 -AW2 8 . (8)
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Il8 STRUCTURAL MECHANICS.
The unknown slopes have thus been eliminated. The next
step is to remove either M or F. Equation (i) must equal M% 9
iorx=li. Therefore
Mi^Mo+Folx-lwxh 2 , or Ff = l J ° +}wi/i.
In the same way, for second span][ F\ =» — j + fad* j
Substitute the values in (8) and obtain
Mpli , Mi/ 2 , M x -M y w x h* M2-M1
22 3 6 6
W2 3 wJi s W2 3
+ ~7^ 8 ^4"'
.\M<fi + 2Mi(h+ti+M&--t(wil l *+wJ#) 9 • • (9)
which is Clapeyron's formula for pier moments for a continuous
beam, with continuous load, uniform per span. Notice the
symmetry of the expression. The negative sign to the second
member indicates that the bending moments at points of sup-
port are usually negative.
Example. — Three spans, 30 ft., 60 ft., and 30 ft. in succession.
Load on first and last 500 lb. per ft., on middle span 300 lb. per ft.
No moment at either outer end. Then Mq—o. M\=M2 by sym-
metry.
2M 1 • 90+ M 2 • 60= — K500 • 3© 3 + 300 • 6o?).
ifi= -81,562* ft.-lb. F = -2,7181+7,500= +4,7*ii lb -
Fi'=-f 4,78i{— 30-500= — 10,218} lb. Fi= 300 -30= 9,000 lb.
/. P =^4=4,78iilb.; Pi = P 2 =io,2i8f+9,ooo= 19,218} lb.
The bending moment and shear at any point can now be readily
determined.
If the two adjacent spans are equal and have the same load,
Mo+4^i+^2=-iw/ 2 (10)
If there are n spans, » — 1 equations can be written between
»+i quantities M , Mi...M n . But if the beam is simply
placed on the points of support, the extremities being unre-
strained, Mq—o and Af n =o, and there remain »-i equations
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RESTRAINED AND CONTIiJUOUS BEAMS. 119
to determine M 1 . . . Af n -i- If the beam is fixed at the ends,
the equations *o*=o and i n =o will complete the required number.
no. Shears and Reactions. — As the shear is the first derivative
of the bending moment, § 56, from (1) is obtained
— -F-Fo^x, (nl
as was to be expected, +F acting upwards on the left of the
section. A similar equation can be written for each, span. fy l*> v
The reaction at any point of support will be equal to the 7 *~
shear on its right plus that on its left with the sign reversed.
As the shear on its left is usually negative, the arithmetical sum
of F n and F n ' commonly gives the reaction.
A simple example may make the application plainer. Given
two equal spans on three supports,
wi=w 2 =w. Mo=o, 3/2=0. (10) gives Afi= — \wP.
F = -\wl+\wl=l wl\ F x ' = lwl-wl= -$wl.
F x = lwl+$wl=iwl; F 2 ' = $wl-wl= -fw/.
Po=$wl; Pi«(f+f)wf-fttrf; P 2 = iwl
1 ivfi
(5) gives 10= -^(o+A-A)^--^/-
(6) gives i\ = -gj(o + A - \)™P = o,
and the analogous equation for the second span is
1 wP
ED ™ T " *' 4&EI'
•which differs from *o only in direction of slope.
(a) gives *T_- _+£«**_.
wP wx*
(4) gives EIv= -~-gx+-frwlx* — — .
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120 STRUCTURAL MECHANICS.
These equations determine the slope and deflection at each
dv
point. Putting T" = o> there results J 3 — 9& 2 + ike 3 — o, containing
the root #=/, already known. Therefore divide by /— x and
obtain P+lx— 8# 2 «=o, which is satisfied for #—0.4215/, the point
of maximum deflection. The substitution of this value in the
equation for v will yield v max%
From (1) M=\wlx—\wx 2 .
If M — o, §/ — £#=0, or #= $/, the point of contraflexure.
Differentiate M and get F=§2f/-2ew.
If F-o, #=§/, the point of +M maXt
Example. — If a uniformly loaded continuous beam covers five
equal spans,
1/0+4^/1+^2= -^2= 3/i+4^2+3f3= ^2+4^3+3/4
-M3+4M4+A/5.
1/0=0: 1/5=0.
Then Mi=-'f s wP=M A ) M 2 =-&wP=Mz.
Po-ilwf-^: Pi-iitrf-^; ft-H**-**
The sum of the reactions must equal $wl.
in. Coefficients for Moments and Shears. — It has been
found that the numerical coefficients for moments and shears
at the points of support, when all spans are equal and the load
is uniform throughout, may be tabulated easily for reference and
use. Thus the values of M and F just obtained for the five equal
spans can be selected from the lines marked V. The reactions
are given by the arithmetical addition of the shears. The sum
of the reactions must equal the total load. The shears at the
two ends of any span differ by the whole load on the span, the
shear at the right end being negative. The dashes represent
the spans.
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RESTRAINED AND CONTINUOUS BEAMS. 121
SHEAR AND REACTION COEFFICIENTS.
I. \-\wl;
n. f-f $-M
in. tV-A tV-A tW*;
iv. \\-hl if-il it— if hh-hb
v. if— ff fs-it it-4| H-B H-tt;
etc., etc.
PIER MOMENT COEFFICIENTS.
II. -i-wP;
III. — I l B — tIj-^hP;
iv. -A-A-A-;
etc., etc.
The rule for writing either table is as follows: For an even
number of spans, the numbers in any horizontal line are obtained
by multiplying the fraction above, in any diagonal row, both
numerator and denominator, by two, and adding the numerator
and denominator of the preceding fraction. Thus, in the first
2X6+ 5 17 . 2X 1 + 1 3
table, — r^ = -^ anc * m the second table, — -5 =-r,
2X10+8 28 2X10+8 28
2X 3+28^ . _ . f . , ,
or w q ; — 5= — . For an odd number of spans, add the two
2X38 + 28 104 r
preceding fractions in the same diagonal row, numerator to
_ 13+ 5 18
numerator and denominator to denominator. Thus, -5- — =-5-.
28 + 10 38
The denominators agree in both tables. A recollection of two
or three quantities will enable one to write all the others.
Example. — Continuous beam of five equal spans, each /, carrying
w per foot. Where and what is the max. + M in second span? Shear
changes sign at \\l from left end of span. If this span were inde-
pendent, +M at that point would be \wl 2 =^- — . The
19-19 3^1 8
negative or subtractive moment is ($V+*Vi$) u ^ 2 - The difference
between these values is + M max -
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122 STRUCTURAL MECHANICS.
A more general investigation will produce equations which
are of great practical value in the solution of problems concerning
continuous bridges, swing-bridges, etc., as follows :
112. Three-moment Theorem for a Single Weight. — O is the
origin, Fig. 55; the supports are at distances h below the axis of
X. A single weight W n is distant kl n from O on the span /„,
k being a fraction, less than unity, of the span in which W is
situated.
The moment at section S beyond W n will be, as in the former
discussion,
M x =Mn+F n x-W n (x-kl n ) (1)
If x=l n , Mz=Mn+i> and from (1)
F n = Mn+ \- Mn + W n (i-k) (ia)
For an unloaded span, W=o } and F m = — — f -.
For the shear on the left of a section at the right end of the
nth span,
Mn+i—Mn
In
F„i'-Fn-Wn- , " -Wnk.
For an unloaded span, W=o, and
/ M m — Af m -i
^m — — ; ■
As F m f is the shear at left of support w, and F m is the shear
at right of the same support ; the reaction there will be the sum
of -F m and -F m ' or
Pm =
M m+l —Mm M m -i—M n
*m *m— 1
To get values of i and v for span l n it is necessary to write
separate equations for the two segments into which W n divides
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RESTRAINED AND CONTINUOUS BEAMS.
"3
the span. Equation (i) applies to tke right segment, and by
omitting the last term the equation for the left segment is obtained.
M (Pv
Equate -gj to ~t— 2 and integrate between the limits of o and
Fig. 05
kl n on the left and between kl n and x on the light. Then on
the left
/kln PkU
dx+F n J o xdx, . . (2)
and on the right
El(P x ~hu) -U.£dx+F.£xdx-W.£(x-kWx. (3)
The sum of (2) and (3) gives the change of slope between
the support n and the section S.
El(^ - *„) = Mnf'dx + F n f'xdx - Wnfjjx - kl n )dx
=M n x+\F n x*-\W n {x-kl n )* (4)
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124 STRUCTURAL MECHANICS.
Integrate the last equation between the same limits as before.
Eliv-te-hJ-lM^+lF^-iWuix-UJ, . (s)
which is the general equation of the curve of the neutral axis,
the term in W disappearing for values of x less than kl n .
If #=/ n , v = &n+i« K the value of F n from (ia) is inserted
in (5), the slope at support » is
in= *n±l^_^ 2M ^ (6)
The equation of the curve is therefore completely deter-
mi ed when M n and M n +\ are known. The equation of this
curve, between W n and the w + ith support is given by (5), and
the tangent of its angle with the axis of X by (4). If the value
of F n from (ia) and of i n from (6) are substituted in (4), and
dv
x=l n > 7- will be the tangent z n +i at n + i,
or i n+l = *=±L^ + ^M n l n + 2M n+1 / n + W n l n *(k -48)].
Remove the origin from O to N, and derive an expression
for i n by diminishing the indices.
Equate with (6) and transpose.
M „_ l/ n - 1 + 2M n (ln-l +/n) + M n + X l n
-WJn^k-sV + P), (7)
which is the most general form of the three-moment theorem
for a girder of constant cross-section. In using the theorem
the following factors may be of service:
k-P = k(i-kXi + k); 2 k-3k 2 + P = k(i-kX2-k).
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RESTRAINED AND CONTINUOUS BEAMS. 125
Pier moments are usually negative and the end moments
zero. When the supports are on a level, hi = h 2 , etc., and the
term containing EI disappears.
Any reaction P n =F n - F n ' ;
.-. Pn= Mn+i r Mn + Mn - i - Mn +w n (x-k) +Wn . l k.
In 'n— 1
It should be borne in mind that all the preceding deflection
formulas are derived on the assumption that the deflection is
due entirely to bending moment, the deformation from shear
being neglected. In a solid beam the amount of deflection due
to shear is very small, but such is not the case with a truss. In
a truss the deflection due to the deformation of the web may in
some cases amount to half the total deflection. For this reason
the deflection formulas and the three-moment theorem should
not be applied to trusses.
Example. — Three-span continuous girder carrying loads as shown.
Supports on a level.
30'
§
I 2o'
§
M
40'
1
40'
2
3
A
50'
A
100'
A
75'
A
2-i5oM 1 +iooM 2 =-2,ooo-2,5oo-|-|-|-i,ooo-io,ooo- T V--^ r -i|
-3,000- 10,000 • 1 V*tV* H-
iooM 1 + 2.i75M 2 =-i,ooo-io,cxx)- 1 V*Tlr , it-3»ooo , io»ooo- 1 ^--^ r 'rt
300M1 + 100^2= — 14,880,000, \ooM x +350^2= — 13,440,000.
M x = - 40,670 ft.-lb. M 2 = - 26,780 ft.-lb.
50 50
-26,780-1-40,670 , 40,670 , 1,000-8 3,000.4 , 2,000-3 , ,^ 1k
"1 = 1 1 r* 1 =* 4-4,152 ID.
100 50 10 10 5
p ^ + 26,780 -40,670+26,780 1,000-2 j 3,000- 6^ a aiSlb
2 75 100 10 10 '
n —26,780 .,
i>a=-^- = -3S7lb.
— 13 + 4,152+2,218 — 357=2,000+1,000+3,000=6,000.
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126 STRUCTURAL MECHANICS.
Examples. — i. A brick wall 16 in. thick, 12 ft. high, and 32 ft.
long, weighing 108 lbs. per cubic ft., is carried on a beam supported
by four columns, one at each end, and one 8 ft. from each end. Find
M at the two middle columns and the reactions.
M =—31,104 ft. -lb.; Pi= 3,024 lb.; P 2 = 24,624 lb.
2. Two successive openings of 8 ft. each are to be spanned. Which
will be stronger for a uniform load, two 8 ft. joists end to end or one
16 ft. long? Find their relative stiffness.
3. A beam of three equal spans carries a single weight. What
will be the reactions and their signs at the third and fourth points
of support when W is in the middle of the first span?
-f f W; +W. ,V
4. A beam loaded with 50 lb. per foot rests on two supports 15 ft.
apart and projects 5 ft. beyond at one end. What additional weight
must be applied to that end to make the beam horizontal at the nearer
point of support? 156^ lb. at the end, or 31 2 J lb. distributed.
5. A beam of two equal spans on level supports carries a single
load, W> in the left span. Prove that
P2=m3*-py>
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CHAPTER VIII.
PIECES UNDER TENSION.
113. Central Pull. — If the resultant tension P acts along the
axis of the piece, the stress may be considered as uniformly dis-
tributed on the cross-section S. If, then, / is the maximum safe
working stress per square inch for the kind of load which causes
P, Fig. 56,
P=/S, or 5=j
for the necessary section which need not be exceeded throughout
that portion of the piece where the above conditions apply.
Changes due to connections will require a larger section.
If owing to lack of uniformity in the material or the direct
application of P at the end of a wide bar to a limited portion
only of the width the stress may not be considered as uniformly
distributed at a particular cross-section, injurious stress may be
prevented by taking the mean stress / at a smaller value and
obtaining a larger cross-section.
If there is lack of homogeneity, or two materials are used
together, or two or more bars work side by side, those fibres
which offer the greatest resistance to stretching will be subject to
the greatest stress. Fortunately the slight yielding and bending
of connecting parts tend to restore equality of action.
A long tension member has a much greater resisting power
against shock than a short one of equal strength per square inch.
See § 20.
114. Eccentric Pull. — If the variation of stress on a cross-
section is due to the fact that the line of action of the applied
127
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128
STRUCTURAL MECHANICS.
force does not traverse the centre of figure of the cross-section
5, the force P' that can be imposed without causing a unit stress
greater than / at any point in the section is less than P of the pre-
ceding formula, and depends upon the perpendicular distance
yo of the action line of P from the centre of 5.
For safe stresses, which must lie well within the elastic limit,
the unit stress is proportional to the stretch, and plane cross-
sections of the bar before the force is applied are assumed to re-
main plane after the bar is stretched. It is impossible to detect
experimentally that this assumption is not true. Were the plane
sections to become even slightly warped, the cumulative warpings
of successive sections in a long bar ought to become apparent to
the eye. No reference is intended here to local distortion preceding
failure.
If the stress on any section is not uniform and the successive
sections remain plane, they must be a little inclined to one another.
The stress on any cross-section 5
must therefore vary uniformly in the
direction of the deviation of the
action line of P f from the centre,
(Fig. 56), and be constant on lines
at right angles to that deviation.
The force, P', acting at a distance,
y , from the axis of the tie is equivalent
to the same force, P', at the centre
and a moment, P'yo- The stress
on each particle of any section may
therefore be divided into two parts, that due to direct -tension and
that due to the moment or to bending. The former is uniform
across the section and is
it-
s'
The latter is zero at the centre of gravity of the section and a
maximum at the edge toward which the load deviates, where it is
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PIECES UNDER TENSION. 1 29
1 M y* _ p 'y<>yi
/-;.H-|(.+ap)-*(.+ap)-
Note that /* is the mean stress, which is always the existing stress
at the centre of gravity of the cross-section.
Example. — A square bar 1 in. in section carries 6,000 lb. ten-
sion. The centre of the eye at the end is i in. out of line. Then
/=6,ooo(i + i*i- 12) = 15,000 lb. per sq. in., 2$ times the mean and
probably the intended stress.
A bar which is not perfectly straight before tension is applied
to it tends to straighten itself under a pull, but the stress will
not become uniform on a cross-section. The bar is weaker
in the ratio of j t to /, as it might carry }S if the force were cen-
tral, but now can safely carry only } t S. If a thrust is applied
to a bent bar, there is a tendency to increased deviation from a
straight line, and to an increase in the variation of stress.
It is seen from the example above that a small deviation yo
will have a decided effect in increasing / for a given P', or in
diminishing the allowable load for a given unit stress. Herein
may be the explanation of some considerable variations of the
strength of apparently similar pieces under test; and, on account
of such effect, added to other reasons, allowable working stresses
may well be and are reduced below what otherwise might be
used.
115. Hooks. — The bending action on and the strength of a
hook are given by the same formulas. Here y\ will be the dis-
tance from the inside edge to the centre of the cross-section, and
yo the distance from the action line of the load to the same centre.
Then the maximum unit tension
,-f (.+*?).
Example. — A hook, the section of which in the bend is elliptical,
iX| in., carries the link of a chain at a distance of J in. horizontally
22 1 • c HP i
from the inside of the bend. Then 5=— -^ = i sq. in.; r*= -7=—,
7 °*4 10 16
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130 STRUCTURAL MECHANICS.
§74, V.; y =i in. Then /=2P(i + i.i-i6)=i8P. If /=8,ooo,
P=45o lb.; if /= 12,000, ^=650 lb. Compare with the given section.
The ordinate of the bend should be reduced as much as possible.
116. Combined Tie and Beam. — If to a tension member
transverse forces are applied, or if it is horizontal and its weight
is of importance, the unit tensile stress on the convex edge, due
to the maximum bending moment, must be added to the unit
stress at that point due to the direct pull. As in § 114,
, {M max ,)y x P
fp = j 1 and ft = £ .
But /=/&+/< must not exceed the safe unit tension, and
the needed section is, since I=Sr 2 f
In this case the sections may vary, since the external bending
moment M varies from point to point.
If the piece is rectangular in section, as with timber, the
formula may be written
, 6M P . 1/6M \
In practical calculation of such a rectangular section, if A is
assumed, it is sufficient to compute the breadth to carry M and
add enough breadth to carry P, when the combined section
will have exactly / at the edge.
Example. — A rectangular wooden beam of 12 ft. span carries a
single weight of 3,000 lb. at the quarter span, and, as part of a truss,
resists a pull of 20,000 lb. If /= 1,000 lb., what should be lhe
section under the weight? Mm^.^— = 81,000 in.-lb.
12
8i,ooo»6 ..« . Tr , , .. 20,000 , _,
— 1 =M 2 =486. If A= 12, 5=3.37. Also, — =1.67. En-
I.OOO I.OOO-I2 '
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PIECES UNDER TENSION. 13 1
tire breadth=3.37+ 1.67 = 5.04. Section=5Xi2 in. The same result
is obtained by the formula
, 1 /6-8i,ooo \
0= ( h 20.000).
I2-I,000\ 12 /
If the tensile stress in a combined tie and beam is small as
compared with the stress due to flexure the above solution is
accurate enough, but when the tensile stress is large the error
is considerable. The transverse forces acting pn the beam
cause it to deflect and consequently the line in which the direct
pull acts does not pass through the centre of gravity of each
section of the beam and a bending moment is produced which
partially counteracts the bending moment due to the transverse
forces. The bending moment at any section then is
yi
in which v is the deflection of the tie-beam at the section where
}b is found. By referring to Chapter VI it is seen that the deflec-
tion of a beam may be written
v = /
where k is a constant depending upon the way in which the beam
is loaded. If it is assumed that v is proportional to /& in the case
under consideration as well as in the case of a beam not subjected
to tension, there results
/»-■
Myi
T ~~JPP'
For a beam supported at both ends and uniformly loaded £= ^;
carrying a single load at the centre, k = -fa\ carrying a single
load at the quarter-point, k= -fa- If the ends are free to turn
and the transverse load is uniformly distributed or is a single
load near the centre, k may be taken as -j^.
An expression for the fibre stress in a horizontal steel eye-bar
bending under its own weight is readily derived from the last
equation. Since a bar of steel one square inch in area and one
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13 2 STRUCTURAL MECHANICS.
foot long weighs 3.4 lb. and E is 29,000,000, there results when
all dimensions are in inches:
4,900,000/i
1 l h Y
/, + 23,000,000(y)
Example. — An eye-bar 8X1 in., 25 ft. long, carries 100,000 lb. ten-
sion. /*= 12,500 lb. per sq. in. If the effect of the deflection is neg-
lected, /&= 2,390 lb. Taking the deflection into account by the last
formula, /&= 1,360 lb.
117. Action Line of P Moved towards the Concave Side. —
It will be economical, if it can be done, in a member having such
compound action, to move the line on which P acts towards the
concave side. If there are bending moments of opposite signs
at different points of the length, or at the same point at different
times, such adjustment cannot be made. If y is made equal
to i(Mmax.) +P> one-half of the bending moment will be annulled
at the point where M maXm exists, and at the point of no bending
moment from transverse forces an equal amount of bending
moment will be introduced. The unit stresses on the extreme
fibres at the two sections will be the same, but reversed one for
the other.
Example. — A horizontal bar, 6X1 in. section and 15 ft. long,
has a tension of 33,750 lb. It carries 100 lb. per foot uniformly dis-
,. ioo-i5-i5-i2 .11
tnbuted. 1/^^.= ^— - — =33,75© in.-Jb. .*. y may be made
i in. Then / from M^.^ 33,750 ' = ± 5,625 lb. on either edge. But
/from P=^|^(i±^) = 5,625(i±i) = 5,625±2,8i2.5. Stress at
top at ends and at bottom at middle 8,43 7$ lb.; at bottom at ends and
at top at middle 2,81 2\ lb. tension.
J The extreme fibre stress from bending moment of the load
varies as the ordinates to a parabola; that from Pyo is constant.
A rectangle superimposed on a parabolic segment will show the
resultant fibre stress at each section.
118. Connecting-rod. — If a bar oscillates laterally rapidly,
as does a connecting-rod on an engine, or a parallel rod on loco-
motive drivers, there are forces developed due to the acceleration,
and at certain positions of the bar these forces are transverse and
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PIECES UNDER TENSION. 133
cause bending. When a parallel rod is in its highest or lowest
position the centrifugal force, due to the circular motion of every
part of the rod, is acting at right angles to the bar, which is then
subjected to a uniformly distributed transverse load of
wv 2 w rt _
— 47m 2 R,
gr g
in which n is the number of revolutions per second, R the radius
of the crank, w the weight of unit length of bar and g the acceler-
ation due to gravity.
The rotating end of a connecting-rod at two points in its path
is under the influence of a transverse force whose intensity is
obtained by the above formula, but as there can be no transverse
force at the sliding end due to acceleration, the rod as a whole
may be considered to be acted upon by transverse forces varying
uniformly from wifi+gr at one end to zero at the other. The
maximum fibre stress due to such loading may be found and
added to the tensile or compressive stress caused by the pull or
thrust along the rod. An I-shaped section is suitable for such
members. Owing to the rapid variations and alternation of stress,
the maximum unit stress should be small. Mass is disadvan-
tageous in such rods.
119. Tension and Torsion. — A tension bar may be subjected
to torsion when it is adjusted by a nut at the end, or by a turn-
buckle. The moment of torsion will give rise to a unit shear
at the extreme fibre, for a round rod, of q\ = T+o.ig6cP by § 82,
or at the middle of the side, for a square rod, of qi = T+o.2o8h 3
by § 83, either of which, combined with /=P-^5, the tensile
stress, will give p x = £/+\/(i/ 2 + ? 2 ). See-§ 86.
Example. — A round bar, 2 in. diameter, to be adjusted to a pull of
10,000 lb. per square inch, calls for the application to the turnbuckle
of 200 lb. with an arm of 30 in., one-half of which moment may be sup-
posed to affect either half of the rod. If the turnbuckle is near one
end, the shorter piece will experience the greater part of the moment.
^ 000 • 2*7
0=- 5-^=1,910 lb. The maximum unit tension on the outside
* 22- I 3
fibres of the rod will be s,ooo+\/(5,ooo 2 -f 1,9102)= 10,350 lb.
120. Tension Connections. — If a tension member is spliced,
or is connected at its ends to other members by rivets, the splice
should be so made or the rivets should be so distributed across
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134 STRUCTURAL MECHANICS.
the section as to secure a uniform distribution of stress. An
angle-iron used in tension should be connected by both flanges
if the whole section is considered to be efficient. One or more
rivet-holes must be deducted in calculating the effective section,
depending on the spacing of the rivets. See § 185. If the stress
is not uniformly distributed on the cross-section, the required
size will be found by § 114, S=-n 1 + 2/-
Transverse bolts and bolt-holes are similar to rivets and rivet-
holes.
Timbers may be spliced by clamps with indents, and by
scarfed joints, in which cases the net section is much reduced;
so that timber, while resisting tension well, is not economical
for ties, on account of the great waste by cutting away. How-
ever, where the tie serves also as a beam, timber may be very
suitable.
121. Screw-threads and Nuts. — If a metal tie is secured by
screw-threads and nuts, the section at the bottom of the thread
should be some 15 per cent, larger than the given tension would
require, to allow for the local weakening caused by cutting the
threads. Bars are often upset or enlarged at the thread to give
the necessary net section and thus save the material which would
be needed for an increase of diameter throughout the length
of the rod.
To avoid stripping the thread, the cylindrical surface, whose
area is the circumference at the bottom of the thread multiplied
by the effective thickness of the nut, should be, when multiplied
by the safe unit shear, at least equal to the net cross-section of
the rod multiplied by the safe unit tension. 2;ri?i-/-ji=;ri?i 2 /,
or fR\ = 2qit. As q\ is usually taken less than /; as with a
square thread only half of the thickness is effective, and with
a standard V thread quite a portion of the thickness must be
deducted, nuts are usually given a thickness nearly or quite
equal to the net diameter of the rod. Heads of bolts may be
materially thinner.
122. Eye-bars. — The eye for the connection of a tension bar
to a pin is seen in Fig. 57. The pin is turned and the eye bored
to a reasonably close fit. Since bearing first takes place at the
back of the pin, the most intense pressure will be found there,
and it will probably diminish at different points of the semi-
circumference until the horizontal diameter is reached. The
pressure on the pin may be found to extend slightly below that
diameter. If these pressures are assumed to be normal, and
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PIECES UNDER TENSION.
'35
are laid off in succession at 2'-2, 2-3, 3-4, . . . 5-6, and closed
with 6-6', the pull on the bar, a pole can be assumed at o and
an equilibrium polygon or curve drawn, cutting the centre line
of the material of the eye about as shown. By moving o hori-
zontally and changing the point of beginning near A vertically, this
equilibrium curve can be brought to the right position to satisfy
the requirement that the sum of the products, from A to D, of
each ordinate normal to the equilibrium curve multiplied by the
force 0-2, 0-3, etc., there acting, shall equal zero. This require-
ment means that the tangents at A and at D to the centre line
shall make the same angle with each other, before and while the
pull is applied. (See Greene's Graphics, Part II., Chap. VI.)
It will be seen that the resultant force 0-1 at the section at A
is smaller than 0-4 or 0-6, the pull at B or C. Hence considerable
deviation of the resultant force
from the centre line at that section
is not a serious matter. The eye
was formerly made with an un-
necessary enlargement at A, but
is now made commonly circular
through more than half of its perim-
eter. The edges of greatest stress
are at A on the outside, B on the
inside, and C on the outside. This
neck should be wide. The material
in front of the pin within the dotted
triangular area is of no service. In
the looped eye of Fig. 57, made by bending the bar around the
pin, that space is empty. Experiment has shown that for strength
this loop should be long, from two to two and one-half diameters
of the pin. If it were not for the weld and the excess of metal
on either side of the pin, such a form of eye would be a satisfactory
one.
If yo is the deviation of the force from the centre of section
at A or B, Fig. 57, the half width being y\ and the pull 0-1 or 0-4
being P, the unit stress at the extreme edge, by § 114, will be
Flff.W
where h = width of eye on one side.
It is necessary to make the pin from § to f the width of the bar,
in order to develop the strength of the latter, that is, to give sufficient
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136 STRUCTURAL MECHANICS.
bearing or compression area back of the pin. The right section
through the eye exceeds that of the bar from 33 to 50 per cent.
Examples. — 1. A round bolt 1 J in. in diameter carries a load of
20,000 lb. As its head is not square to its length, the centre of resist-
ance is probably J in. from the axis of the bolt. What is / in this case,
and how much greater than the mean stress?
/=4i,5oo lb. Since the elastic limit has been passed the actual
maximum stress is probably less.
2. Find the maximum stress and the mean stress for a pull of
20,000 lb. on a square eye-bar i$Xi$ in. if the pin is $ in. out of cen-
tre. 26,670 lb.
3. A rectangular bar, section 2X1 in., has a central pull of 8,000
lb. Then /= 4,000 lb. If the bar is widened on one side to 3 in. with-
out change of force and its point of application, what is /? 5,333 lb.
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CHAPTER IX.
COMPRESSION PIECES— COLUMNS, POSTS, AND STRUTS.
123. Blocks in Compression. — If the height of the piece is
quite small as compared with either of its transverse dimensions,
and the load upon it is centrally imposed, the load or force P may
reasonably be considered as uniformly distributed over the cross-
section 5, and the unit stress / upon each square inch of section
will be given by the formula
P-/S, or /=|,
as is the case with any tension member when the force is centrally
applied.
124. Load not Central. — So also when the action line of
the resultant load cannot be considered as central, but deviates
from the axis of the piece a distance yo, the force P can be replaced
by the same force acting in the axis and a couple or moment Py 0>
which moment must be resisted at every cross-section by a
uniformly varying stress, forming a resisting moment exactly
like that found at a section of a beam. Compare Fig. 56, and
change tension to compression.
If } c is the uniform unit stress due to a central load of P, and
if jb is the unit stress on the extreme fibre lying in the direction
in which yo is measured, the latter stress being due to the moment
Pyo,
* -L. t _ p ypyi_ p ypvi
137
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138 STRUCTURAL MECHANICS.
The greatest unit stress on the section is
/-/.+/.-f(-^)-4^)-
The load that such a piece will carry is
is
r
2
By comparison with the formula of the preceding section
it will be seen that the piece, when the load is eccentric, is weaker
in the ratio
The values of y\S+I or yi-v-r 2 are given below for some
of the common sections of columns, y\ being measured in the
direction h.
I y x S y x +r*
Rectangle — \h oh r-
Square — \h h 2 r-
Circle ^ id frd* I
04 a
btf-Vh* 17 L , L „, 6h(bh-b'h')
Hollow rectangle. . . }h bh — b'h' -77^ — , /y/Q -
12 bhr—bh*
Hollow circle ^"^ # ^((f 2 -^ 2 ) j^^
125. The Middle Third. — The mean or average unit stress
is always found at the centre of gravity of the cross-section.
On sections symmetrical about the neutral axis the fibre stress
can be zero at one edge only when the fibre stress on the oppo-
site edge is twice the mean, that is, when
.2
lb=f e or y = -
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COMPRESSION PIECES. 139
This condition is satisfied for a rectangle when yo = \h. Hence for
a rectangular section in masonry the centre of pressure must
not deviate from the centre of figure more than one-sixth of
the breadth in either direction, if the unit stress at the more
remote edge is not to be allowed to become zero or tension. As
masonry joints are supposed in many cases not to be subjected
to tension in any part, the above statement is equivalent to say-
ing that the centre of pressure or line of the resultant thrust
must always lie within the middle third of any joint.
Likewise, for two cylindrical blocks in end contact, the centre
of pressure should fall within the middle fourth of the diameter
if the pressure is assumed to be uniformly varying.and it is not
permissible to have the joint tend either to open or to carry ten-
sion at the farther edge.
The unit pressure at the most pressed edge of a rectangle
can be found for any deviation yo-
1st. When the stress is over the whole joint, as before,
/
-&+¥)■
2d. When compression alone is possible, and only a part
of the surface of the joint is under stress. The distance from
the most pressed edge to the action line of P is JA— yo- The
entire pressed area=3(£&— yo)b, since the ordinates representing
stresses make a wedge whose length along y is three times the
distance of P from the most pressed edge.
P=i/-3(i*-yo)&; }=SP+tth-y )b.
If the case is that of a wall, and P is the resultant force per
unit of length, 6 = 1.
As yo increases, / increases, until finally the stone crushes
at the edge of the joint, or shears on an oblique plane as described
in § 23. Sometimes the pressure is not well distributed, from
poor bedding of the stones, and spalls, or chips, under the action
of the shearing above referred to, may break off along the edge,
without failure being imminent, since when the high spots break
off others come into bearing.
P can never traverse one edge of the joint, if tension is not
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140 STRUCTURAL MECHANICS.
possible at the other edge, as the unit stress then becomes infinite.
Some writers commit an error in determining the thickness of
a wall by equating the moment of the overturning force about
the front edge or toe with the moment of the weight of the wall
about the same point. This process is equivalent to making
the action line of P traverse that point. The centre of moments
should be taken either at the outer edge of the middle third of
the joint, when pressure is desired over the whole joint, or about
a point at such a distance £A— yo from the front as will give
maximum safe pressure at the front edge. A uniformly vary-
ing stress extending over three times that distance will equal P,
as lately stated. A portion of the joint at the rear will then
tend to open.
Examples. — A short, hollow, cylindrical column, 12 in. external
diameter, 10 in. internal diameter, supports a beam which crosses the
column 2 in. from its centre.
Sd 8- 12
& /-£(■*£)-"'*•*
d?+d' 2 144+100 244'
or the stress at either edge will be 80% greater and less than the mean
stress.
A joint, 10 ft. broad, of a retaining-wall is cut at a point 3 ft. 9 in.
from the front edge by the line of the resultant thrust above that joint.
If this thrust per foot of length of the wall is 28,000 lb., the pressure per
square foot at the front edge will be — ! — (i-fii-^ (r ) = i|- 2,800=4,900
lb. At the rear it will be 700 lb. per sq. ft.
126. Resistance of Columns. — A column, strut, or other
piece, subjected to longitudinal pressure, is shortened by the
compression. As perfect homogeneousness does not exist in
any material, the longitudinal elements will yield in different
amounts, so that there is apt to be a slight, an imperceptible
tendency to curvature of the strut. Hence the action line of
the applied load may not traverse the centres of all cross-sec-
tions of the piece. The product of the applied force into the
perpendicular distance of its line of action from the centre of
any cross-section will be a bending moment, which must develop
a resisting moment at the cross-section, resulting in a varying
stress, as in § 124. This curvature under longitudinal pressure
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COLUMNS, POSTS, AND STRUTS. 141
can be readily obtained with test specimens of most materials,
even with some samples of cast iron, and the form of the curve
apparently conforms to the one to be deduced by theory. A
tendency to such a curve must therefore exist under working
stresses, although the curvature is imperceptible, unless the
column happens to have its load perfectly axial, a contingence
that cannot be safely relied upon. The column formula, so
called, should therefore be confidently applied.
Further, as such curvature can be produced in test speci-
mens not more than four or five diameters long, such a formula
is applicable to columns and struts of any length. It is not
necessarily to be applied, however, to very short posts, or blocks,
for the relation P=/5 will determine their size with sufficient
exactness.
127. The Yield-point Marks the Column Strength. — The
influence of yo in determining the load a compression piece
will carry has been shown in § 124 to be very marked. A column
which has become sensibly bent under a load is very near com-
plete failure. The moment of the load at the cross-section of
greatest lateral deflection has then become so large that the
stress on the extreme fibres passes the yield-point, and the great
increase of deformation at and above the yield-point at once
increases the bending moment greatly. Hence it is true that
the yield-point marks nearly the ultimate compressive strength
of materials when tested in column form.
Again, the fact that, in tests of large columns, a very slight
shifting of the point of application of the load at either end has
a decided influence on the amount of weight such a column
will carry is a confirmation of the statements with which this
discussion opened. It also has a bearing upon the truth of the
theoretical deduction as to the effect of eccentric loading, as
discussed in § 124, and to be applied to long columns later.
128. Direction of Flexure. — The flexure usually occurs,
unless there is some defect or weakness, in a direction parallel
to the least transverse dimension of the strut, i.e., perpendicular
to that axis in the cross-section which offers the least resisting
moment. By the application of longitudinal pressure to a slender
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142 STRUCTURAL MECHANICS.
rod its flexure may be made very apparent. The form of the
column formula ought to resemble that of § 124,
129. The Ideal Column. Euler's Formula. — Assume the
column to be perfectly straight and homogeneous and the load
to be applied axially. Under such ideal conditions the load can
cause no flexure, but if a small deflection is given the column
by the application of a lateral force, it is desired to find what
load is just sufficient to maintain that deflection after the lateral
force is removed.
If the column is fixed in direction at its ends, by its connections
to other pieces, or by having a broad, well-bedded base and cap,
it will act in flexure much as a beam fixed at the ends.
p ] j A couple or bending moment, which may be rep e-
sented by Mo, will thus be introduced at each end.
Let P= applied external force or load; ?/=any deflec-
tion ordinate, measured at right angles to the action
i ( , x f ti ne °f P> fro m the original axis of the column to any
! point in the axis when bent; x= distance from one
end along the original axis to any ordinate v; /= length
of column.
The combination of the moment Mq at the end
of the column with the force P has the effect, as shown
in mechanics, of shifting the force P laterally a distance
Mq +P=Vq) hence the action line of P is now parallel
to the original axis, at a distance v from it, or in the
line F E of Fig. 58. The ordinate to the points of
p | t contraflexure is therefore v . This action can be
I r more fully realized by conceiving that the bearing
surface at A is removed, and that P acts at such a point
on a horizontal lever as to keep the tangent to the curve at A
strictly vertical.
As v is measured from the original axis, the bending moment
at any section is M = P(vq-v), which will change sign when the
second term is larger than the first.
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COLUMNS, POSTS, AND STRUTS. 1 43
If the flexure is very slight, an equation similar to that used
with beams may be written
dv
Multiply by -7- and integrate.
EI
2
(£) 2=p(v ° r -^ )+c -
dv dv
As t~=o when v=o, C=o. When T~=o at J?, the middle
of the length, Vw^ =2^0, or double the ordinate at the point of
contraflexure, F. Extracting the square root of the above equation
and separating the variables gives
dx. - dv
p V2V0V—V 2 '
V
D versin -1 — +C
As v=o when#=o, C'=o.
i/=v versin (^nJ^j) = *>o(i -cos (^Yg/))-
As 1 —cos 0=2 sin 2 £0,
v=2v sin 2 \$x s lgj}.
If, in this equation, x=\l a value of v,^. is obtained to be
equated with the previous value, 2V .
Vmax. = 2^0 = 2V sin 2 (i^NJ ^j) •
sin [Vsl^j) - 1 ; WfiJ ™ **•
4**£J
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144 STRUCTURAL MECHANICS.
which is commonly known as Enter's formula. P is the ultimate
strength of the column and is independent of the deflection; that
is, within certain limits P is the same, no matter what small
deflection is given the column. Under a load less than P the
column would straighten; under a greater load it would fail.
The equation of the curve of the column is
v = 2V sin 2
(r)-
To find the points of contraflexure, make v=*Vq.
sin vjx) =\/£ =:S in 45° — sin —
7C
4
x
.*. j = J or x = \l from either end.
Hence the curve is made of four equal portions, A F, F D,
D E, and E H.
A column hinged, pin-ended, or free to turn at its ends, and
of length represented by EF = £/, will have the same portion
of stress at D that is due to bending as does a column of length
AH=/, which is fixed at its ends.
If, in actual cases, F is considered to be practically in the
same position horizontally as before loading, it may be said
that a column fixed at one end and hinged at the other, of length
FH = J/, will also have the same portion of stress arising from
bending. The maximum deflection* will then occur at one-third
of its length from the hinged end. This result has been verified
by direct experiment on a full-sized steel bridge member.
130. Rankine's Column Formula. — Euler's formula can be
put in the form
P y?E
S~//\ 2 '
&
P I
This curve, when plotted with 77- and — as variables, has
•j r
the form of the curve B C, Fig. 59, where the greatest mean unit
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COLUMNS, POSTS, AND STRUTS.
145
stress which any ideal column of a given ratio of / to r can endure
is given by the ordinate to the curve. But as the yield-point
marks the ultimate strength of metal in compression, for short
columns the line A B, whose ordinate is the yield-point, takes
the place of the curve.
For long, slender columns . (when l+r exceeds 200, say)
Euler's formula gives results close to the ultimate strength found
by loading actual columns to destruction; but for shorter columns
the formula gives results much too large. It must be remembered
that actual columns do not satisfy the conditions from which
Euler's formula was derived. No real column is perfectly straight
A
\b
30000-
onnnrv
yO
P _
s
1,144,(XX
),000
ffl
2
40,000
"Tfi"
, c
28,600 r
100
800
900
400
Fig. 59
and homogeneous, nor can the load be applied exactly along
the axis, and as no two columns are exactly alike no theoretical
formula can exactly fit all cases. For these reasons Euler's
formula is never used in designing.
When the results of actual column tests are plotted, they
follow in a general way some such curve as ADC. Rankings
P p
formula, -=-= ~-, is the equation of such a curve, in which
r 2
p is the yield-point of the material and a is a numerical coefficient.
By modifying the value of a the curve can be made to agree
fairly well with the results of tests. For a short block when
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146 STRUCTURAL MECHANICS.
l+r approaches zero the formula becomes P+S=p. If a is
made equal to p -4-4^ 2 £, the curve of Rankine's formula approaches
Euler's for large values of /-*-r, where Eider's agrees well with
the results of tests. This value of a has been used by some
authorities and is believed to give conservative values. The
curve of Fig. 59 was so plotted. However, the value of any
column formula depends upon its agreement with the results
of tests, and in practice a is generally an empirical factor.
In order that P of Rankine's formula may be the load which
an actual column can safely carry instead of its ultimate strength,
the greatest allowable unit stress on the material, /, is to be sub-
stituted for the yield-point, p. The formula then becomes
P I
S l 2 '
It is to be regarded as giving the mean unit stress on any cross-
section when the stress on the extreme fibre at the point of greatest
deflection is /.
In designing columns the load, P, and the length, /, are usu-
ally known. The form of cross-section is then assumed which
fixes S and r. If the assumed values satisfy the equation the
section chosen is satisfactory; if not, another trial must be made.
The radius of gyration, r, to be used is the one perpendicular
to the neutral plane, that is, it is measured in the direction in
which the column is likely to deflect. Unless the column is
restrained in some way it will, of course, deflect in the direction
of the least radius of gyration of the cross-section.
Example. — What load will a 12-in. 3i$-lb. I beam 10 ft. long carry
/ P \
as a column if the mean unit stress is 12,000-4- ( i-f — -^ J ? r about
axis through web =1.01 in. 5=9.26 sq. in. -g = — ! = 14,100.
^^=0.392. i^2=8,62o lb. ^=8,620X9.26= 79,800 lb.
36,000 Jy 1.392 y /y '
131. Multipliers of a. — As seen above, 2/ should theoretically
be substituted for / when the column is hinged or free to turn
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COLUMNS, POSTS, AND STRUTS. 1 47
at its ends, in order to obtain the equivalent length of a column
which is fixed at the ends; and for a column fixed at one end
and hinged at the other $/ should be substituted for I for the
same reason. Or, more conveniently, a may be used for a column
fixed at the ends and of length /; \a for a column hinged at
both ends and of length /; and tya for a column hinged at one
end and fixed at the other, length /. Actual tests, carried, how-
ever, to the extreme of bending or crippling, appear to show
that a column bearing on a pin at each end is not hinged or per-
fectly free to turn; hence the multipliers of a more commonly
used, instead of i, */, and 4, are 1, f, and 2.
132. Pin Friction. — Some regard columns as neither per-
fectly fixed nor perfectly hinged, and use but one value of a
for all, which might perhaps then be taken as a mean value.
The moment of friction on a pin is considerable. If P is the
load on a post or strut, d the diameter of the pin, and tan <f> the
coefficient of friction of the post on the pin, the moment of fric-
tion at the pin will be P -\d- tan <f>\ and this moment, if greater
than Mo=Pvo, will keep the post restrained at the end, so that
the tangent there to the curve remains in its original direction.
As P and Vq increase, Mo will become the greater when Vo exceeds
\d tan <f>\ the column will then be imperfectly restrained at its
ends, and the inclination will change. As the friction of motion
is less than that of rest, such movement when started may be
rapid. Some tested columns, showing at first the curve of Fig. 58
known as triple flexure, have suddenly sprung into a single curve
and at once offered less resistance.
It may be doubted whether ordinary columns under working
loads ever develop a value of Vq sufficient to overcome the pin
friction, unless the column is very slender and the diameter
of the pin small. The custom is quite general of using the same
column formula for struts with fixed and with hinged ends.
133. Straight-line Column Formula. — In engineering struc-
tures columns which have a greater ratio of length to radius
of gyration than 100 or 150 are very rarely used. Within this
limit a straight line can be drawn which will represent the aver-
age results of a series of column tests, plotted as described in
§ 130, as well as Rankine's formula does. As the equation of a
straight line is easier to solve than the equation of a curve, the
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148
STRUCTURAL MECHANICS
For working values it has
straight-line formula is much used,
the form
P # l
in which / is the greatest allowable unit stress on the material
and c is a numerical coefficient determined empirically. In this
country both Rankine's and the straight-line formulas are exten-
sively used in designing columns. Examples of each will be
found in Chapter X.
134. Swelled Columns. — Some posts and struts, especially
such as are built up of angles connected with lacing, are swelled
or made of greater depth in the middle. If the strut is perfectly
free to turn at the ends, such increase in the value of r 2 may be
quite effective, and r 2 for the middle section may be used in
determining the value of P, provided the latter does not too
closely approach the uniform compression value at the narrower
ends. But if the strut is fixed at the ends, or is attached by a pin
whose diameter is large enough to make a considerable friction-
moment, such enlargement at the middle is useless ; for an equally
large value of r 2 ought to be found at the ends also. Hence
swelled columns and struts are but little used.
Vo>
Fig. 60
135. Column Eccentrically Loaded*
— When the load is applied eccen-
trically to a long column, the maxi-
mum unit stress found in the extreme
fibre on the concave side must be due
to three combined effects :
1st. The stress due to the load P 9
OTp =P+S. (Fig. 60.)
2d. The stress due to the resisting
moment set up by Pyo.
3d. The stress due to the resisting
moment set up by Pvq.
From§ I24 , /^( I+ M»)- A , + A ,^l.
From § 130, / = j {1 + a-^J =p + p aj r
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COLUMNS, POSTS, AND STRUTS. 149
If then the column is long and the line of action of P deviates
from the original axis of the column by a distance yo> the three
expressions, po, poyoyi +**> and p aP +r 2 should be added, giving
p , Wi\ . P f
This formula may be put in a form more convenient for
designing:
o P Pyoyi
p
The first term gives the area of cross-section necessary to
resist the direct thrust and the second term the area to resist the
bending moment due to eccentric loading.
Since y will determine the direction of flexure, r must be
taken in this case in the direction yo; that is, the moment of
inertia and r must be obtained about the axis through the centre
of gravity and lying in the plane of the section, perpendicular to
yo- That the moment Py > although small, has a decided weaken-
ing effect on a column is proved by experiment, and its unintended
presence may explain some anomalies in tests.
136. Transverse Force on a Column. — The resisting power
of a column or strut to which a transverse force is applied in ad-
dition to the load in the direction of the axis, and the proper
dimensions of the strut, are involved in some doubt. Theo-
retically, the formula is deduced as follows:
From the formula for the resisting moment of a beam,
M = fl+yi, the stress on the outer layer from such bending
moment is My\ -*-/. Hence if M is that particular value of the
bending moment (from the transverse load or force) which exists
at the section of maximum strut deflection, where the column stress
is greatest (that is, at the middle for a column with hinged ends, but
perhaps at the ends for a column with fixed ends, since M may
then be greater at the ends), the maximum unit stress on the
concave side
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ISO STRUCTURAL MECHANICS.
.Pi P\ Myi
p
S P~
c P ...M*
p
If the straight-line formula is used instead of Rankine's* die
last equation has the form
' f
Or, again, it may be said that at the section in question P is
moved laterally by the moment M a distance yo=M +P. Then,
by § i35>
P_ f
S I 2 MyS
which reduces to the forms given above.
The value of r 2 to be used in these formulas is that for flexure
in the plane of M .
Example. — A deck railway -bridge with 20-ft. panels has ties laid
directly on the top chord, thus imposing a load of 2,500 lb. per foot. If
the direct thrust is 249,000 lb., what should be the size of chord for a
working stress on columns of 10,000— 45/ ^r and on beams of 9,000 lb.
per sq. in. ? Try a section composed of two 20-in. 65-lb. I beams and
one 24X£-in. cover-plate. 5=50.16 sq. in. The centre of gravity of
the section lies 2.45 in. above middle of I beams. y\ to upper fibre =
8.05 in. 7=3,298 in. 4 r about horizontal axis =8. 11 in. Jf=
i- 2,500- 20-20-12= 1,500,000 in.-lb. Working column stress= 10,000
— 45X240-7-8.10=8,670 lb. per sq. in. 249,000-7-8,670=28.7 sq. in.
. , , . . 1,500,000X8.05 . . .
required for column action. — — Q = 20.4 sq. m. required
9,000X0.11 Xo.II
for beam action. 28.7+20.4=49.1 sq. in., total. Assumed section is
sufficient.
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COLUMNS, POSTS, AND STRUTS. 15 1
137. Lacing-bars. — The parts of built-up posts are usually
connected with lacing straps or bars. These bars carry the
shear due to the bending moment arising from the tendency of
the post to bend and should be able to stand the tension and
compression induced by the shear. At the ends and where
other members are connected, in order to insure a distribution
of load over both members, batten or connection plates at least
as long as the transverse distance between rivet rows are used
in good practice.
Each piece should be of equal strength throughout all its
details. A post or strut composed of two channels connected
by lacing-bars and tie-plates is proportioned for a certain load,
the mean unit stress being reduced in accordance with the formula
in which the variable is the ratio of the length to the least radius
of gyration of the whole section. In the lengths between the
lacing-bars, this ratio for one channel with its own least radius
should not be greater than for the entire post. Nor should the
flange of the channel have any greater tendency to buckle than
should one channel by itself. The same thing applies to the
ends of posts, where flanges are sometimes cut away to admit
other members. Quite a large bending moment may be thrown
on such ends, when the plane of a lateral system of bracing does
not pass through the pins or points of connection of the main
truss system.
The usual bridge specification reads : Single lacing-bars shall
have a thickness of not less than *V, and double bars, connected
by a rivet at their intersection, of not less than ^V of the distance
between the rivets connecting them to the member. Their
width shall be: For 15-in. channels, or built sections with
3J- and 4-in. angles, 2 \ in., with f-in. rivets; for 12- or 10-in.
channels, or built sections with 3-in. angles, 2j-in., with f-in.
rivets; for 9- or 8-in. channels, or built sections with 2j-in. angles,
2 in., with f -in. rivets.
All segments of compression members, connected by lacing
only, shall have ties or batten plates placed as near the ends as
practicable. These plates shall have a length of not less than
the greatest depth or width of the member and shall have a thick-
ness of not less than -jV of the distance between the rivets con-
necting them to the compression member.
Examples. — 1. A single angle-iron 6X4Xf in. and 6 ft. 8 in. long
is in compression. Use r=o.g or obtain it from § 76. 5=3.61 sq. in.
If P-r-S= 12,000— 34/ -s-r, what will it carry? 32,400 lb.
2. A square timber post 16 ft. long is expected to support 80,000 lb.
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1 52 STRUCTURAL MECHANICS.
If /—i,ooo and the subtractive term is 2/4- r, what is the size?
10X10 in.
3. What is the safe load on a hollow cylindrical cast-iron column
13 ft. 6 in. long, 6 in. external diameter, and 1 in. thick, if it has a broad,
flat base, but is not restrained at its upper end? (/= 9,000 lb., E=
17,000,000, 5=15.7 sq. in.) 124,000 lb.
4. If a short wooden post 12 in. square carries 28,800 lb. load,
and the centre of pressure is 3 in. perpendicularly from the middle of
one edge, what will be the maximum and the mean unit pressure, and
the maximum unit tension, if any ? 500 lb. ; 200 lb. ; — 100 lb.
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CHAPTER X.
SAFE WORKING STRESSES.
138. Endurance of Metals Under Stress. — It is important to
determine how long a piece may be expected to endure stress
when constant, when repeated, or when varied and perhaps
reversed; and it is still more important to find what working
stresses may be allowed upon a given material in order that
rupture by the stresses may be postponed indefinitely.
The experiments carried on by Wohler and Spangenberg, and
afterwards continued by Bauschinger, show the action of iron
and steel under repeated stress. Tests were made on speci-
mens in tension, bending, and torsion. * A number of bars of
wrought iron and steel were subjected repeatedly to a load which
varied between zero and an amount somewhat less than the ulti-
mate strength. When the bar broke under the load a similar
bar was tested under a reduced load and was found to resist a
greater number of applications before rupture. Finally a load
was reached which did not cause rupture after many million
repetitions. The stress caused by such load was taken as the
safe strength of the material and was called the primitive safe
strength The following table is an illustration:
WOHLERS TESTS OF PHCENIX IRON IN TENSION.
Stress varying between Number of Applications.
o and 46,500 lb. per sq. in 800
o " 43,000 " " " " 107,000
o " 39,000 " " " " 341,000
o " 35,000 " " " " 481,000
o " 31,000 " " " " 10,1 2,000
* For a description of machines and tests, see Unwin, Testing of Materials of
Construction.
f 53
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154
STRUCTURAL MECHANICS.
As a result of Wohler's and Spangenberg's experiments the
latter states that alternations of stress may take place between
the following limits of tensile (— ) and compressive ( + ) stress in
pounds per square inch with equal security against rupture:
PHCENIX IRON AXLE.
— 15,500 +15,500 Difference 31,000
—29,000 o " 29,000
-43,000 -23,500 " i9>5°°
KRUPP CAST-STEEL AXLE.
—27,000 +27,000 Difference 54,000
—47,000 o " 47,000
— 78,000 —34,000 " 44,000
UNHARDENED SPRING STEEL.
-48,500
— 68,000
— 78,000
-87,500
Difference 48,500
* * 44,000
11 39>ooo
1 ' 29,000
o
— 24,000
-39,000
-58,50°
The figures are approximate.
The results illustrate Wohler's law, that rupture 0} material
may be caused by repeated alternations 0) stress none 0) which attains
the absolute breaking limit. The greater the range 0} stress the
smaller the stress which will cause rupture.
The term jatigue 0} metals is often applied to the phenomena
just described.
139. Bauschinger's Experiments. — Professor Bauschinger,
besides making tests similar to Wohler's, determined the effect
which repeated stresses had upon the position of the elastic
limit and the yield -point. (§13, 14, and 15.) He derived the
following results from tensile tests:
Ultimate
Strength, /.
Wrought -iron plate
Bessemer soft-steel plate.
Iron, flat
Iron, flat
Thomas steel axle
Primitive
Safe
Stress, u.
Original
Elastic
Limit.
Original
Yield-
point.
28,000
14,800
29,700
34,000
33>9°0
41,800
31,000
25,700
32,600
34,000
32>3°°
35-IOO
43,000
38,100
54,600
62,000
57,600
57. 200
87,000
X
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SAFE WORKING STRESSES 155
From an inspection of the table it is seen that the primitive
safe stress on sound bars of iron and steel for tensions alternating
from zero to that stress is a little less than the original yield-point
of the metal Although the safe stress was found to be greater
than the original elastic limit, yet when bars which remained
unbroken after several million applications of stress were tested
statically it was found that the elastic limit had risen above
the stress applied. The inference is that the primitive safe
stress is below the elastic limit although the elastic limit may
be an artificially raised one.
Pulling a bar with loads above the original elastic limit and
below the yield-point raises the elastic limit in tension but lowers
it in compression and vice versa. These artificially produced
elastic limits are very unstable. When a bar is subjected to a
few alternations of equal and opposite stresses which are equal
to or somewhat exceed the original elastic limits, the latter tend
toward fixed positions called by Bauschinger natural elastic
limits. He advanced the view that the original elastic limits
of rolled steel and iron are artificially raised by the stresses set
up during manufacture. These natural elastic limits seem to
correspond with Wohler's range of stress for unlimited alternating
stresses.
140. Alteration of Structure. — If a bar has been subjected to
repetitions of stress somewhat below the yield-point, any general
weakening or deterioration in the quality of the bar ought to
show itself in some way; but bars which have endured millions
of repeated stresses or bars which have broken under repeated
stresses give no indication of any weakening when tested statically.
Instead of being weakened by repeated stresses well below the
primitive safe strength, a bar is really improved in tenacity.
Bars fractured in the Wohler machines did not draw out,
no matter how ductile the material might be, but broke as if
the material were brittle. From these facts we are forced to
conclude that whatever weakness there may be is confined to
the surface of rupture. Such breaks appear to be detail breaks.
Continuity is first destroyed at a flaw or overstressed spot, and
from that point the fracture spreads gradually until the section
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156 STRUCTURAL MECHANICS.
is so weakened that the load is eccentric and bending stresses
are set up which cause the piece to break by flexure.
From the results of Wohler's and Bauschinger's tests it would
appear that steel might undergo an indefinite number of repe-
titions of stress within the elastic limit without rupture, or that
within that limit the elasticity was perfect. However, certain
experiments on the thermoelectric and magnetic properties of
iron under stress and on the molecular friction of torsional pen-
dulums give results inconsistent with the theory of perfect elas-
ticity. Engineers differ more or less in their interpretations
of the experimental results but agree in using lower working stresses
for varying than for static loads.
141. Launhardt-Weyrauch Formula. — A number of formulas
based on Wohler's experiments have been advanced for the
determination of the allowable unit stress on iron and steel when
the range, as well as the magnitude, of the stress is considered.
Launhardt proposed the following formula for the breaking
load of a member which is subjected to stresses varying between
a maximum and a minimum stress of the same kind:
a=ul
/— wmin. stressX
u max. stress/ "
a = breaking load under the given conditions.
w = primitive safe stress.
/ = ultimate strength, static.
v= greatest stress piece will bear if repeated from +v to — v
indefinitely.
Weyrauch extended Launhardt's formula to cover cases of
alternate tension and compression, in which case the minimum
stress is to be considered as negative.
/ u— v min. stress\
\ u max. stress/
As the results of the different experiments vary more or less,
the different authorities do not agree as to the values of /, w,
and v to be substituted in the formulas. In some of Wohler's
experiments u appears to be greater than \t and to approxi-
mate to §/. This value is assumed by Weyrauch and gives
(/— w)-i-tt=$. Likewise if v = \u the coefficient of the second
formula becomes \ and both formulas reduce to
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SAFE WORKING STRESSES. 1 57
=*(i+i
mm. stress \
max. stress/'
the sign of the second term changing for reversed stresses. The
choice of values of u and v to use in the equations seems to have
been determined largely by the result sought. The formulas
are of rather doubtful value, but they are more or less used for
determining working stresses in steel bridges.
The following relationships of /, w, and v are given by some
authorities, and they conform more nearly with the results of the
experiments quoted in this chapter:
Steady load a=t
Load varying from o to u a=w = \t
11 " " +7/to-7/ a=v = J/
142. Reduction of Unit Stresses. — The safe working stress to
be used for any material will depend upon several considerations :
Whether the structure is to be temporary or permanent; whether
the load is stationary or variable and moving; if moving, whether
its application is accompanied by severe dynamic shock and
perhaps pounding; whether the load assumed for calculation is
the absolute maximum; whether such maximum is applied rarely
or is likely to occur frequently; whether the stresses obtained
are exact or approximate; whether there are likely to be secondary
stresses due to moments arising from changes of the assumed
frame; what the importance of the piece is in the structure, and
the possible damage that might be caused by its failure.
The allowable unit stresses of different kinds, and for greater
or less change of load, will be further reduced to provide against :
Distribution of stress on any cross-section somewhat different
from that assumed; variations in quality of material; imperfec-
tions of workmanship, causing unequal distribution of stress;
scantness of dimensions; corrosion, wear, or other deterioration
from lapse of time or neglect; lack of exactness of calculation.
The allowable unit stresses so determined will be but a small
fraction of the ultimate or breaking strength of the material; and
it is evident that the idea that it will require several times the
allowable maximum working load to cause a structure to fail is
seriously in error.
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158 STRUCTURAL MECHANICS.
Overconfidence of the in experienced designer in the correct-
ness of his design may be checked by a study of this section.
143, Load and Impact. — The design should be completely
carried out, both in the principal parts and in the details. The
latter require the most careful study, that they may be at once
effective, simple, and practical.* All the exterior forces which
may possibly act upon a structure should be considered, and due
provision should be made for resisting them. The static load,
the live load, pressure from wind and snow, vibration, shock, and
centrifugal force should be provided for, as should also deteriora-
tion from time, neglect, or probable abuse. A truss over a machine-
shop may at some time be used for supporting a heavy weight by
block and tackle, or a line of shafting may be added; a small
surplus of material in the original design will then prove of value.
Light, slender members in a bqdge truss, while theoretically
able to resist the load shown by the strain sheet, are of small
value in time of accident. The tendency from year to year is
towards heavier construction.
Secondary stresses, as they are called, are due — first, to the
moments at intersections or joints, when the axes of the mem-
bers coming together at a connection do not intersect at a common
point ; and second, to the moments set up at joints by the resistance
to rotation experienced by the several parts when the frame or
truss is deflected by a moving load. If symmetrical sections
are used for the members, if the connecting rivets are symmetrically
placed, and if the axes of the intersecting members meet at one
point, secondary stresses will be much reduced.
All members of a structure should be of equal strengh, and
the connections should develop the full strength of the body of
the members connected. The connections should be as direct
as possible. When a live load is joined to a static load, the
judgment of the designer, or of the one who prepares the specifi-
cations for the designer, must be exercised. A warehouse floor
to be loaded with a certain class of goods has maximum stresses
from a static load. The floor of a drill-room, ball-room, or
highway bridge receives maximum loading from a crowd of people
* A portion of these paragraphs is extracted from a lecture by Mr. C. C. Schneider.
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SAFE WORKING STRESSES. 159
the possible density of which can be ascertained. But if these
masses of people keep moving, and more particularly if they keep
step, the effort of their weight will be increased by the vibrations
resulting therefrom. This action is generally called impact.
In the case of a building, the floor- joists, receiving the impact
directly, will be most affected; the girders which carry the joists
will be less affected, and the columns which support the girders
will receive a smaller percentage of the impact, the proportionate
effect growing less as the number of stories below the given floor
increases. In the absence of trustworthy data from which to
determine this impact, it is left to the judgment of the engineer
to increase the live load by a certain percentage, or to decrease
the allowable unit stress, for each case, to provide for the effect,
as will be seen in the values given later.
For economy, make designs which will simplify the shop work,
reduce the cost, and insure ease of fitting and erection. Avoid
an excess of blacksmith work and much use of bent pieces.
144. Dead and Live Load. — From what has been said it is
readily seen that a moving or live load has a much more serious
effect on a truss than a static or dead load, and in the case of a
railroad bridge the stresses in the members due to a rapidly
moving train are much greater than the stresses would be under
such a train at rest. In designing the members of steel bridges
this increase of stress is usually provided for by one of three ways:
Using different working stresses for dead and live load, as a
unit stress of 18,000 lb. per sq. in. for dead load and 9,000 lb. for
live. When this method is employed the ratio is commonly 2 to 1.
Using a varying unit stress found from a modified form of the
Launhardt-Weyrauch formula, as /=o,ooo(i4 : ). If
J ' y \ max. stress/
the stresses reverse the fraction is negative. The coefficient J in
the original formula is dropped arbitrarily.
Adding a certain percentage of the live load, called impact,
to the sum of the dead- and live-load stresses, and using a constant
working stress as 18,000 lb., one impact formula often used is
300
Impact = jn — —7- X live-load stress. By "length" is meant
the distance in feet through which the load must pass to produce
maximum stress in the member in question. Such experiments
as have been made on bridges under moving load indicate that
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i6o
STRUCTURAL MECHANICS.
the impact is not as great as the formula gives. The third method
is sometimes combined with the first or second, but with the use
of a different impact formula.
While these three ways of designing are quite different in theory
and method, the resulting sizes of members are much the same.
145 , Working Stresses for Timber. — The following table is
from Bulletin No. 12, U. S. Dept. of Agriculture, Div. of Forestry:
Safe Unit Stresses at 18% Moisture.
Structures freely exposed to the weather.
«
Compression.
Bending.
Shear.
With
Grain.
Across
Grain.
Extreme
fibre.
With
Grain.
Tension.
Long-leaf pine
Short-leaf pine
White pine
1,000
84c
700
760
880
650
800
215
215
150
140
170
"5
400
i»55°
1,300
880
1,090
1,320
1,200
"5
IOO
75
200
12,000
9,000
7,000
Norway pine
Douglas spruce
Redwood
White oak
10,000
Factor of safety
5
3
5
4
1
The values given were found by taking the average of the
lowest 10 per cent, of the results of tests on non-defective timber
and dividing by the factor of safety given in the last row. As
timber never fails in tension the safe tensile stress is not given;
the last column shows the ultimate strength. For structures
protected from the weather the compressive and bending values
may be increased 20 per cent. The shearing value should not
be increased. The table was constructed for designing railway
trestles.
The following safe working stresses have been recommended
by the Committee on Strength of Bridge and Trestle Timbers,
Am. Assn. of Ry. Supts. of Bridges and Buildings :
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SAFE WORKING STRESSES.
161
Tension.
Compression.
Bend-
ing.
Shear.
With
Grain.
Across
Grain.
With Grain.
Across
Grain.
Ex-
treme
Fibre.
With
Grain.
End
Bear-
ing.
Cols,
under
Diams.
Across
Grain.
Long-leaf pine
White pine.
1,200
700
800
1,200
800
600
700
1,000
60
50
1,600
1,100
1,200
1,600
1,200
1,000
700
800
1,200
800
800
8co
900
350
200
200
300
200
150
2CO
500
1200
700
700
1,100
700
600
750
1,000
I50
100
I50
100
100
100
200
1,250
Soo
Norway pine
Douglas spruce
Spruce and Eastern fir .
Hemlock
750
600
Redwood
White oak
200
1,400
1,000
Factor of safety
10
10
5
5
4
6
4
4
Messrs. Kidwell and Moore deduced the following column
formulas based on the values given in the preceding table. Least
cross-dimension of column = h; length =/.
l+h<2$ l+h>2$
Long-leaf pine 1,000— 8.5/ -4-A 1,000— nl + h
White pine 700 -6. / -4- h 700 - 8/ +h
Norway pine, spruce, and
Eastern fir 800-6.5/ -f-A 800- 8l+h
Douglas spruce 1,200—9. l + h 1,200— i2l+h
Hemlock 800 — 7.6/-5-A 800— 9.5/^A
Redwood 800— 8.6/ -s- h 800 — nl+h
Oak 900—9. l+h 900— nl+h
For buildings Mr. C. C. Schneider recommends the following
unit stresses :
Compression.
Bending.
With Grain.
Across
Grain.
Extreme
Fibre.
Shear with
Grain.
End
Bearing.
Columns
under 10
Diameters.
Long-leaf pine
White pine, spruce. . .
Hemlock
1,500
1,000
800
1,200
1,000
600
500
1,000
350
200
200
500
1,500
1,000
800
1,200
100
100
100
White oak
200
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1 6a STRUCTURAL MECHANICS.
P } I
For columns when /-*-/*> 10, -£■ = { 7-, in which / is
o 100 h '
the compressive stress for a short column taken from the table.
The strength of columns built up by bolting two or more
pieces of timber together is no greater than that of the individual
pieces acting independently.
The names of wo9ds vary in different localities; long-leaf,
short-leaf, and one or two other Southern species of pine are not
distinguished in the market and are called Southern, Georgia, or
yellow pine. Douglas spruce is also known as Oregon fir, spruce,
or pine.
146. Working Stresses for Railroad Bridges. — Mr. Theodore
Cooper's Specifications, edition of i^oi, give the following unit
stresses in pounds per square inch.
MEDIUM STEEL IN TENSION.
Main truss members — dead load 20,000
live " 10,000
Floor-beams and stringers 10,000
Lateral and sway bracing — wind load 18,000
" " " live " 12,000
Floor-beam hangers and members liable to sudden loading 6,000
For soft steel in tension reduce the unit stresses 10 per cent.
MEDIUM STEEL IN COMPRESSION.
Chord segments — dead load 20,000 — 90/ -fr
" " live " 10,000 -45/ -=-r
Posts of through-bridges — dead load 1 7,000—90/ -s-r
" " " live " 8,500-45/^r
Posts of deck-bridges and trestles — dead load 18,000 —80/ -s-r
" " " " " live ".... 9,000-40/^-r
Lateral and sway bracing — wind load 13,000 — 6o/ + r
li live " 8, 700 -40/ + r
tt a
For soft steel in compression reduce the unit stresses 15 per
cent.
The ratio of / to r shall not exceed 100 for main members and
120 for laterals.
When a member is subjected to alternating stresses, each
stress shall be increased by eight-tenths of the lesser and the
member designed for that stress which gives the larger section.
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SAFE WORKING STRESSES. 1 63
PINS AND RIVETS.
Shear 9,000
Bearing (thickness X diameter) 1 5,000
Bending (pins) 18,000
For floor rivets increase number of rivets 25%.
" field " " " " " 50%.
' lateral " decrease " " " 33%.
PEDESTALS.
Pressure on rollers, pounds per lineal inch, 30oXdiam. in
inches.
Pressure of bedplates on masonry, pounds per square inch,
250.
The specifications of the American Bridge Co., 1900, give the
following working stresses:
The maximum stress in a member is found by adding the dead
300
load, live load, and impact. Impact = — yXlive-load stress,
in which L is length in feet of distance to be loaded to produce
maximum stress in member.
Soft Steel. Medium SteeL
Tension 15,000 17,000
„ . 15,000 17,000
Compression ^— —
i3»5 c or 2 n,ooor 2
Shear on web-plates 9,000 10,000
The ratio of / to r shall not exceed 100 for main compression
members and 120 for laterals.
When a member is subjected to alternating stresses the total
section shall be equal to the sum of the areas required for each
stress.
PINS AND RIVETS.
Shear 11,000 12,000
Bearing (thickness Xdiameter) 22,000 24,000
Bending (pins) 22,000 25,000
For hand-driven field rivets increase mimber 25%.
" power-driven " " " " 10%.
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1 64 STRUCTURAL MECHANICS.
PEDESTALS.
Pressure on rollers, pounds per lineal inch,
i2,oooVdiam. in inches.
Pressure of bedplates on masonry, pounds per square inch,
400.
147. Working Stresses for Highway Bridges. — Mr. Cooper's
Specifications, 1901, give the following stresses:
MEDIUM STEEL IN TENSION.
Main truss members — dead load 25,000
" " " live " 12,500
Floor- beams, stringers, and riveted girders 13,000
Lateral and sway bracing — wind and live load 18,000
Floor-beam hangers and members liable to sudden loading 8,000
For sojt steel in tension reduce the unit stresses 10 per cent.
MEDIUM STEEL IN COMPRESSION.
Chord segments — dead load 24,000 — 1 10I -j-r
" " live " 12,500-- 55/^r
Posts of through -bridges — dead load 20,000 — 90/ -4-r
" " " live " 10,000- 45/ -rr
Posts of deck-bridges — dead load 22,000 — 80/ -r-r
" {" " live " 11,000- 40/ -*-r
Lateral and sway bracing — wind load 13,000 — 60/ +r
11 " " live " 8,700- 4 o/-r
For soft steel in compression reduce the unit stresses 15 per
cent.
The ratio of / to r shall not exceed 100 for main members and
120 for laterals.
When a member is subjected to alternating stresses, each
stress shall be increased by eight-tenths of the lesser and the
member designed for that stress which gives the larger section.
PINS AND RIVETS.
Shear 10,006
Bearing (diameter X thickness) 18,000
Bending (pins) 20,000
For floor rivets increase number 25%.
" field " " " 5°%-
" lateral " decrease " 30%.
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SAFE WORKING STRESSES. 165
PEDESTALS
Pressure on rollers, pounds per lineal inch, 3ooXdiam. in
inches.
Pressure of bed-plates on masonry, pounds per square inch,
250.
TIMBER FLOOR- JOISTS.
Fibre stress on yellow pine or white oak 1,200
" " " white pine or spruce 1,000
The working stresses given by the American Bridge Co.
Specifications for Highway Bridges, 1901, are the same as for
Railway Bridges, with the following exceptions:
The sum of dead and live load is to be increased by 25 per
cent, of the live load to compensate for the effect of impact and
vibration.
The ratio of / to r shall not exceed 120 for main members and
140 for laterals.
When a member is subjected to alternating stresses, design
for that stress which gives the larger section.
Fibre stresses on floor- joists same as in preceding specification.
148. Working Stresses for Buildings. — A set of building specifi-
cations by Mr. C. C. Schneider may be found in the Transactions
of the American Society of Civil Engineers, Vol. LIV, June,
1905. Some of the working stresses recommended are abstracted
below.
Structural steel of ultimate strength between 55,000 and
65,000 lb. per sq. in. is to be used.
Tension 16,000
Compression 16,000 — 70/ -s-r
Shear on web plates (gross section) 10,000
For wind bracing and the combined stresses due to wind and
the other loading the permissible stresses may be increased 25
per cent.
The ratio of / to r shall not exceed 125 for main compression
members and 150 for wind bracing.
Members subjected to alternating stresses shall be propor-
tioned for the stress giving the larger section.
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166 STRUCTURAL MECHANICS.
PINS AND RIVETS.
Shear on pins and rivets 12,000
" " bolts 9,000
Bearing on pins and rivets 24,000
" bolts 18,000
Bending on pins 24,000
For field rivets increase number 33 per cent.
WALL PLATES AND PEDESTALS.
Bearing pressures in pounds per square inch on
Brickwork, cement mortar 200
Rubble masonry, cement mortar 200
Portland cement concrete 350
First-class masonry, sandstone 400
" " " granite 600
MASONRY.
Permissible pressure in masonry, tons per square foot :
Common brick, natural cement mortar 10
" i ' Portland cement mortar 12
Hard-burned brick, Portland cement mortar 15
Rubble masonry, Portland cement mortar 10
Coursed rubble " " " 12
Portland cement concrete, 1-2-5 2 °
149. Machinery, etc. — The designing of machinery has not
been systematized as has the designing of structural steel work,
and the choice of working stresses is to a large extent a matter of
individual judgment. The following values are taken from a
table in Unwin's Machine Design and are for unvarying stress;
if the stress varies, but does not reverse, multiply by two-thirds;
if the stress reverses, multiply by one-third.
Tension.
Cast iron 4,200
Iron forging 15,000
Mild steel 20,000
Cast steel 24,000
For shafting of marine engines 9,000 lb. per sq. in., on wrought
iron and 12,000 lb. on steel is commonly used, the shaft being
designed for the maximum and not the mean twisting moment.
These values are about those used in other machinery.
The working stress in boilers made of 60,000 lb. steel may
be as high as 13,000 lb. per sq. in. with first-class workmanship,
but ordinarily 10,000 to 11,000 lb. is employed.
ompression.
Torsion.
12,600
2,IOO
15,000
6,000
20,000
8,000
24,000
10,000
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CHAPTER XI.
INTERNAL STRESS: CHANGE OF FORM.
150. Introduction. — Let any body to which external forces
are applied be cut by a plane of section. The force with which
one part of the body acts against the other part of the body is the
stress on the plane of section. This stress is distributed over the
section and may be uniform or may vary. Its intensity, that is,
I he unit stress, at any point is found by dividing the stress on an
indefinitely small area surrounding the point by the area. Stresses
may be normal, tangential or oblique to the surface on which they
act, and they are completely determined only when both intensity
and direction are known. It is desirable to know the magnitude
and kind of the unit stress at each point in order to be sure that
the material can safely resist it; or to determine the required
cross-section to reduce the existing stresses to safe values.
A unit stress is expressed as a certain number of pounds of
tension, compression, or shear on a square inch of section. If
the plane of section is changed in direction, the force on the
section may be changed and the area of section may also be changed
so that the unit stress on the new section is altered from that on
;he old in two ways. Stresses per square inch, or unit stresses,
therefore cannot be resolved and compounded as can forces,
unless they happen to act on the same plane. Generally, each
unit stress may be multiplied by the area over which it acts, and
the several forces so obtained may be compounded or resolved
as desired; the final force or forces divided by the areas on which
they act will give the desired unit stresses.
Where the stress on a plane varies from point to point, as
does the direct stress on the right section of the team, and as does
the shearing stress also in the same case, the investigation is
supposed to be confined to so small a portion of the body that
the stress over any plane may be considered to be sensibly constant.
167
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168
STRUCTURAL MECHANICS
151. Stress on a Section Oblique to a Given Force. — Suppose
a short column or bar to carry a force of direct compression or
tension, of magnitude P, cen-
trally applied and uniformly
distributed over the cross-
section, S. The unit stress on
and perpendicular to the right
section will be pi = P+S.
On an oblique section CD,
Fig. 2, making an angle
with the right section AB,
the unit stress will be P+S
sec = pi cos 0, making an
angle of with the normal to the oblique section on which
it acts. If this oblique unit stress is resolved normally and
tangentially to C D, the
Normal unit stress =p n = pi cos 0* cos 0=pi cos 2 6;
Tangential do. = q =p\ cos sin 0.
The normal unit stress on the oblique plane is of the same
kind as P, tension or compression; the tangential unit stress, or
shear, tends to make one part of the prism slide down and the
other part slide up the plane.
The largest normal unit stress for different planes is found
when 0=o, which defines the fracturing plane for tension; the
minimum normal unit stress occurs for = go°; and the greatest
unit shear is found for = 45°, when we have qmax.^bpi-
152. Combined Stresses. — The action line of P may be taken
for the axis of X. Two equal and opposite forces, pull or thrust,
may then be applied along the axis of Y, and the normal and
tangential unit stresses found on the plane just discussed; and
similarly for the direction Z. The normal unit stresses, since
they act on the same area, may then be added algebraically, and
the shearing stresses may be combined; finally a resultant ob-
lique unit stress may be found on the given plane.
Thus, if pi is the unit stress on a section normal to the X
axis, and p2 the unit stress on a section normal to the Y axis,
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INTERNAL STRESS: CHANGE OF FORM.
169
pn = p\ cos 2 + p 2 cos 2 (9o° - 0) = p x cos 2 + p 2 sin 2 0,
j =^1 sin cos 0—^2 cos sin = (^1 —^2) sin cos 0.
^ n is a maximum when 0=o or 90 , and q is a maximum when
0=45°.
Pnmax. = Pl 0T /> 2 ;
9max.= = h(pl-p2)'
A more convenient method will, however, be developed and
used in the following sections. As most of the forces which act
on engineering structures lie in one plane or parallel planes, such
cases chiefly will be considered.
153. Unit Shears on Planes at Right Angles. — If, in the pre-
ceding illustration, the unit stresses, both normal and tangential,
are found on another plane N which makes an angle , = go a —d
with the right section, there will result
pn = pi cos 2 0' = pi sin 2 ; q* - q.
Hence, on a pair of planes of section, both of which are at right
angles to the plane of external forces and to each other, the unit
shears are 0} equal magnitude, and, since pn+pn^pi, the unit
normal stresses are together equal to the original normal unit
stress. It is further evident that one normal unit stress p n ' may
be found by subtraction as soon as the other is known, and that
ordinary resolution on these two planes of the original unit stress
would be erroneous.
154. Unit Shears on Planes at Right Angles Always Equal. —
Since, as before stated, other forces, in other directions, may be
simultaneously applied to the given a
body, and their effects found on the 1 *x
same two planes, it follows that, in p t*p\
any body under stress, equal unit
shearing stresses will exist on two "*~~
planes each of which is perpendicular
to the direction of the shear on the
other. J
Example. — A closed cylindrical re- Fi«. 61
ceiver, J in. thick, has a spiral riveted
joint making an angle of 30 with the axis of the cylinder, and
Kttjtt
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170 STRUCTURAL MECHANICS.
a portion 2 in.X4 in. of the cylinder, Fig. 61, has the given tensions of
2,500 lb. acting upon it. Then ^1 = 2,500-4-2- J= 5,000 lb. per sq. in.,
and ^2— 2,500-7-4- J= 2,500 lb. per sq. in.
pn**S,ooo-t+ 2,500-1=4,375 lb. per sq. in.,
^=5,000-0.433-2,500-0.433 = 1,082 lb. per sq. in.,
^=V(4,375 2 +i,o82 2 )=4,507 lb. per sq. in.,
or 4,507-^=1,127 lb. per linear inch of joint, which value will deter-
mine the necessary pitch of the rivets for strength.
The stress on a joint at right angles to the above can be similarly
found. An easier process will be given in § 161.
155. Compound Stress is the internal state of stress in a body
caused by the combined action of two or more simple stresses
(or balanced sets of external forces) in dLTerent directions, as in
the above example. The investigations which follow are those
of compound stress, but they will, as above stated, be chiefly
confined to stresses in or parallel to one plane.
156. Conjugate Stresses : Principal Stresses. — If the stress on
a given plane in a body is in a given direction, the stress on a
plane parallel to that direction must be parallel
Jf*" to the first-mentioned plane. For the equal re-
i/ sultant forces exerted by the other parts of the
IT*
M
A
°y
*
body on the faces A B and C D of the pris-
matic particle, Fig. 62, are directly opposed to
/C\ * one another, their common line of action trav-
Fig 62 ersing the axis of X through O ; and they are
therefore independently balanced. Therefore
the forces exerted by the other parts of the body on the faces A D
and B C of this prism must be independently balanced and
have their resulcants directly opposed; which cannot be unless
their direction is parallel to the plane Y O Y.
A pair of stresses, each acting on a plane parallel to the direc-
tion of the other, is said to be conjugate. Their unit values are
independent of each other, and they may be of the same or oppo-
site kinds. If they are normal to their planes, and hence at
right angles to each other, they are called principal stresses.
Examples. — The unit stress found in § 154 makes an angle with the
plane on which it acts whose tangent is 4,375-1-1,082 = 4.04. Upon a
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INTERNAL STRESS: CHANGE OF FORM.
171
new plane cutting the metal in this direction the stress must act in a
direction parallel to the joint referred to.
If a plane be conceived parallel to a side-hill surface, at a given
vertical distance below the same, the pressure at all points of that
plane, being due to the weight of the prism of earth above any square
foot of the plane, will be vertical and uniform. Then must the pressure
on a vertical plane transverse to the slope be parallel to the surface of
the ground. That the pressure against the vertical plane is not hori-
zontal, but inclined in the direction stated, is shown by the movement
of sewer trench sheeting and braces, when the braces are not inclined
up hill, but are put in horizontally.
157. Shearing Stress. — If the stresses on a pair of planes are
entirely tangential to those planes, the unit shears must be equal.
Consider them as acting along the
faces of a small prismatic particle
ABCD, which lies at O. The
moment of the total shear on the
two faces A B and C D must bal-
ance the moment for the faces A D
and B C, for equilibrium.
-**
Fig. 63
jABEF^ADHG.
But the area of A B C D, ABEF=AD HG; .\ tf = q.
This construction shows further that a shear cannot act alone
as a simple stress, but must be combined with an equal unit shear
on a different plane.
158. Two Equal and Like Principal Stresses. — If a pair of
principal stresses, § 156, are equal unit stresses of the same kind,
pi and P2, Fig. 64, the stress on every plane is normal to that
plane, and of the same kind and magnitude.
1 w
a"
o
;P'
u:
&
p.
°'t
Fig. 64
Let Pi act in the direction O X on the plane O' B' of the
prismatic particle O' A' B' which lies at O, and p% act in the direc-
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172 STRUCTURAL MECHANICS.
tion Y on the plane O' A', p x and p 2 being equal unit stresses
of the same kind. Make OD^ r O'B', the total force on
O'B' , and O E=/> 2 -0' A', the total force on O' A', both being
positive. Complete the rectangle O D R E. Then must R O,
applied to the plane A'B', be necessary to insure equilibrium
of the prism O' A' B'. Hence /=RO -*-A' B'. Since p x =/>*
OP OE OR. . «_ h _ h
O' B'~0' A'~A' B" " F -r 1 -? 2 -
Because of similarity of triangles A' O' B' and O E R, R O is
perpendicular to A B, or f to A' B', and is of the same kind as
Pi and ^2*
Example. — Fluid pressure is normal to every plane passing through
a given point, and equal to the pressure per square inch on the horizontal
plane traversing the point. Here manifestly the three coordinate axes
of X, Y, and Z might be taken in any position, as all stresses are princi-
pal stresses.
159. Two Equal and Unlike Principal Stresses. — If a pair of
principal stresses are equal unit stresses of opposite kinds, as p\
and - p2, Fig. 64, the unit stress on every plane will be the same
in magnitude, but the angle which its direction makes with the
normal to its plane will be bisected by the axis of principal stress,
and its kind will agree with that of the principal stress to which
it lies the nearer.
In this case lay off Oe in the negative direction, to represent
—p2-0" A"; construct the rectangle O Dre, and draw rO which
will be the required force distributed over A" B" to balance the
forces O" B" and O" A". This force rO will be the same in
magnitude as RO, making f=pi=p2 and rO will make the
same angle with OXorOYasRO does. As R O lies on the
normal to A B, and O X bisects R Or, the statement as to position
is proved. The stress f/ agrees in kind with that one of the
principal stresses to which its direction is nearer.
160. Two Shears at Right Angles Equivalent to an Equal Pull
and Thrust— If the plane A" B" is at an angle of 45 with O X,
rO will coincide with A B and becomes a shear. Therefore two
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INTERNAL STRESS: CHANGE OF FORM. 173
equal unit stresses of opposite kinds, that is a pull and a thrust
and normal to planes at right angles to one another, are equivalent
to, and give rise to equal unit shears on planes making 45 with
the first planes and hence at
right angles to each other; and
vice versa.
Example. — If, at a point in the
web of a plate girder, Fig. 65, there
is a unit shear, and nothing but
shear, on a vertical plane, of 4,000 ^ " p . ^
lb., there must be a unit shear of
4,000 lb., and nothing but shear, on the horizontal plane at that point
and on the two planes inclined at 45 to the vertical through the same
point there will be, on one, only a normal unit tension of 4,000 lb., and
on the other an equal normal unit thrust.
161. Stress on any Plane, when the Principal Stresses are
Given. — This general problem is solved by dividing it into two
special cases; the one that of two equal and like principal stresses,
the other that of two equal and unlike principal stresses. Let
the two principal unit stresses be p\ = D, and ^=OF, of any
magnitude, and of the same kind or opposite kinds. Fig. 66.
The direction, magnitude and kind of the unit stress on any plane
A B is desired.
Let pi be the greater. Divide pi and p2 into their half sum
and difference as follows :
Pi = l(pi+p2)+i(pi-p2), and p2 = \{pi+p2)-\(pi-p2).
The distance O C or O E will represent the half sum
h(Pi+p2), and CD or EF the half difference \(p\-p2)- If
pi and p2 are the same sign the right-hand figure will result;
if of opposite signs, the left-hand figure will be obtained. -
By the principle of § 158, when the two equal principal unit
stresses O C and O E are considered, lay off O M on the normal
to the plane whose trace is A B, for the direction and magnitude
of the unit stress on A B due to $(^1+^2). There remain
C D and E F representing + \(p\-p2) on the vertical axis,
and —\(Pi~~Pz) on ^e horizontal axis respectively.
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174
STRUCTURAL MECHANICS.
By § 159, lay off O Q, making the same angle with O X as
does O M, but on the opposite side, or in the contrary direction,
for the magnitude and direction of stress on plane A B due to
± J(^i— p2)< As O M and O Q both act on the same unit of
area of A B, R O, in the opposite direction to their resultant
O R, will give the direction and magnitude of the unit stress on
A' B' to balance pi on O' B' and p 2 on O' A'. In the figure
on the right R O is positive, or compression. If, in the figure
on the left, where pi is 4- and p2 is — , R O falls so far to the left
as to come on the other side of A B, it will agree with p 2 , and
Fig. 66
be negative or tension. If A B is taken much more steeply
inclined, such will be the case. The small prisms illustrate the
constructions. If R O falls on A B, it will be shear. Some
constructions for different inclinations of plane A B will be help-
ful to an understanding of the matter.
A much abbreviated construction is as follows: — Strike a
semicircle from M on the normal, with a radius MO = $(pi+p 2 ),
and draw M R through the points where the semicircle cuts the
axes of pi and p 2 . The angle N M R is thus double the angle
MOD, since it is an exterior angle at the vertex of an isosceles
triangle. Lay off MR = |(^ -/> 2 ) in the direction of the axis of
greatest stress, and R O will be the desired unit stress on A B.
If p 2 is opposite in kind to pi, M R will be greater than M O,
and R will go beyond P.
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INTERNAL STRESS: CHANGE OF FORM. 175
162. Ellipse of Stress. — For different planes A B through
0, pi and p2 being given, the locus of M is a circle of radius
%(pi+p2)> and the locus of R is an ellipse (as will be proved
below), with major and minor semidiameters pi and p2- Hence
it is seen why pi and p2, normal to the respective planes, and at
right angles, are called principal stresses.
If three principal stresses, coinciding in direction with the
rectangular axes of X, Y and Z, simultaneously act on a given
point, an ellipsoid constructed on them as semidiameters will
limit and determine all possible stresses on the various planes
which can be passed through that point in the body.
That the locus of R is an ellipse may be proved as follows:
Drop a perpendicular R S from R on O X. P M O and O M G
are isosceles triangles. <POM=<MPO=0.
OM = MP = £(/>i+/>2); GR=/>i; PR=/> 2 .
R S =PRsinMPO = /> 2 sin0,
S O = GRcosMPO=/>!Cos0,
OR=p r =V(S0 2 +RS)=V(pi 2 cos 2 d+p2 2 sin 2 d), . (1)
which is the value of the radius vector of an ellipse in terms of
the eccentric angle, the origin being at the centre.
As <NMR = 2<POM = 2 0,
sinMORisin OMR = RM:OR = i(ft-i'2):^
/. sinMOR=sin2 0.^^ ? , (2)
which gives the obliquity of the unit stress to the normal to the
plane, in terms of the angle of the normal with the axis of greater
principal stress, or of the plane with the other axis. The graphical
construction gives the stress and its angle with the normal or
the plane by direct measurement, and is far more convenient
and less liable to error.
If pi=p 2 , the case reduces to that of § 158 or § 159.
If the ellipse whose principal semidiameters are p x and p2
is given, the unit stress on any plane may briefly be found by
drawing the normal to the plane, laying off OM = J(^+^),
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X
176 STRUCTURAL MECHANICS.
taking a radius of \{P\ —p2)> and, with M as a centre, cutting
the ellipse at R on the side of the normal towards the greater
axis. A line R O will be the desired unit stress.
Example. — Let £1 = 100 lb. on sq. in., p2= — 50 lb. Plane AB
makes 30 with direction of p2, or its normal makes 30 with pi. Con-
struct the figure and find the magnitude, direction, and sign of the unit
stress on the oblique plane. Try other values.
163. Shearing Planes. — To determine the angle of the planes
on which there is only shear, and the conditions which render
shearing planes possible.
If the plane A B of the previous figure is to be the shearing
plane, there must be no normal component upon it, and therefore
from § 151, if the plane makes an angle 0, with p2, or the normal
to it is inclined at an angle 0, to pi,
pn = Pi cos 2 0, + p 2 sin 2 0, =0.
COS0, >» p2
No shearing plane is possible unless Pi and P2 differ in sign.
There will then be two planes of shear making equal angles with
the direction of p2 or of pi.
In the above example, V(ioo -*- 50) =\/2 =tan 0, = 1.4142.
= 54° 44'.
If the ellipse of stress is drawn, take a radius equal to the
side of a right angled triangle whose other side is \{P\+p2), and
hypothenuse is \(pi—p2), and strike a circle from the centre of
the ellipse. Planes drawn through the points of intersection of
this circle and ellipse and the centre will be the shearing planes.
Unless pi and p2 differ in sign, the circle will be imaginary. The
value of the shear on these planes is
q = V(l(Pl-p2) 2 -l(pl+p2) 2 )=^ ~plp2-
164. Given any Two Stresses: to Find Principal Stresses. —
As, in actual practice, two oblique unit stresses on different planes
may often be known in magnitude, direction and sign, it will be
required to find the principal unit stresses, since one of them is
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INTERNAL STRESS: CHANGE OF FORM.
177
the maximum stress to be found on any plane, and the other is
the minimum stress of the same kind, or the maximum normal
stress of the opposite kind.
Given two existing unit stresses, p and f of any direction, mag-
nitude and sign, to determine the principal unit stresses, pi and p2.
If pi and p2 were known, and p and f were then to be found
from the former, the construction shown in Fig. 67 would be
made, in which OM = OM'=|(ft+^ and MR=M'R'
s =h(Pi—p2)' If one of these normals were revolved around O
to coincide with the other, the point M' would fall on M, but
M' R' would diverge from M R, while equal in length to it.
Hence, when p and f are the given quantities, let A B, Fig.
68, represent the trace of the plane on which p acts, O N the
Fig. 67
normal to that plane, and O R the unit stress p in magnitude
and actual direction of action on A B. OR represents either
tension or compression, as the case may be. Now let the plane
on which f acts, together with its normal and f/ itself in its relative
position, be revolved about O until it coincides with A B. Its
normal will fall on O N and p f will be found at O R', on one
side or the other of O N, if it is of the same kind with p\ or it
is to be laid off on the dotted line below, if of the opposite sign.
In other words, lay off f from O, at the same angle with O N
which it makes with the normal to its own plane. It is well, for
accuracy of construction, to draw it on the same side of the
normal as p, the result being the same as if it were drawn on
the other side. (The change from one side of the normal to
the other simply consists in using the corresponding line on the
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178 STRUCTURAL MECHANICS.
other side of the main axis of the ellipse of stress). Thus is
found O R/ or — O R' as the case may be. Draw R R' and, since
R M R' must be an isosceles triangle, bisect R R' at T and drop
a perpendicular to R R' from T on to the normal, cutting the
latter at M. Then since, as previously pointed out, OM =
KP1+P2) and MR = MR' = J(^i~^ 2 ),— with M as a c-ntre,
and radius M R, describe a semicirc e; ON will be pi and O S
will be p2. Since p is in its true position, and the angle NMR=»
2 M O D of Fig. 66 or 2 M O X of Fig. 67, the direction of the
axis X along which pi acts will bisect N M R, and the axis along
which p2 acts will be perpendicular to axis X. They may be now
at once drawn through O, if desired.
165. From any Two Stresses to Find Other Stresses. — From
the preceding construction, § 164, the stress on any other plane
may now be found. All possible values of p consistent with the
two, O R and O R', first given, will terminate, in Fig. 68, on the
semicircle just drawn, as at R", and the greatest possible ob-
liquity to the normal to any plane through O will be found by
drawing from O a tangent to this semicircle.
166. When Shearing Planes are Possible. — In case the lower
end of the semicircle cuts below O, Fig. 69, pi and p2 are of
opposite signs, all obliquities of stress are possible, and the dis-
tance from O to the point where the semicircle cuts A B, being
perpendicular to the normal O N, gives the unit
shear on the shearing planes. If pi and p2 are
drawn through O in position, and the ellipse of
stress is then constructed on them as semi-
diameters (as can be readily done by drawing
two concentric circles with pi and p2 respectively
as radii, and projecting at right angles, parallel
to pi and p2, to an intersection, the two points where any radius
cuts both circles), an arc described from O, with a radius equal
to this unit shear and cutting the ellipse, will locate a point in
the shearing plane which may then be drawn through that point
and O. Two shearing planes are thus given, as was proved to
be necessary, § 157.
The above solution may be considered the general case.
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INTERNAL STRESS: CHANCE OF FORM.
179
167. From Conjugate Stresses to Find Principal Stresses. —
If p and f/ are conjugate stresses, it is evident, from, definition,
and from Fig. 62, that they are equally inclined to their respective
normals. Hence O R' will fall on O R, when revolved, both O R
and O R' lying above O when of the same sign, and on opposite
sides of A B when of opposite signs. The rest of the construction
follows as before, being somewhat simplified.
It may be noted that, when pi = ± p2, the propositions of
§§158 and 159 are again illustrated.
One who is interested in a mathematical discussion of this
subject is referred to Rankine's "Applied Mechanics," where it
is treated at considerable length. This graphical discussion is
much simpler, less liable to error, and determines the stresses in
their true places.
168. Stresses in a Beam — The varying tensile and compres-
sive stresses on any section of a beam are accompanied by vary-
ing shearing stresses on that section and by equal shears on a
longitudinal section. The direct or normal stresses due to bend-
ing moment vary uniformly from the neutral axis either way,
§ 62; the shears are most intense at the neutral axis, § 72. The
normal and shearing stresses on the cross-section of a rectangular
beam and also the resultant stresses on the section are shown in
\ oc
-<
1
i*-
/ a
i
I
r*
>
K
/
\
rO'
w *
-n-
Flg.70
Fig. 70, a and b. The maximum and minimum unit stresses
(that is, the principal stresses) at any point, with their directions,
may be found by a slight modification of § 164 as follows :
Let Fig. 71 represent the stresses acting on a particle, as the
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i8o
STRUCTURAL MECHANICS.
one at O, Fig. 70. Lay off O N=/> n on the normal to the plane
A B, on which it acts in its true position. As q acts on the same
Fig. 71
area of A B as does p H9 lay off R N-y at right angler,. The two
component stresses being R N and N O, a line from R to O will
represent the direction and magnitude of the unit stress on A B.
Revolve the horizontal plane through 90 to coincide with A B.
Then lay off R' O = q as shown. (As explained in § 164, R' O may
be laid off on the other side of the normal, O N, but that construc-
tion is not quite so simple.) Since R O and R' O represent p and
f of the preceding sections, connect R and R' and bisect R R' at
M, which point falls on O N and is also the point where the per-
pendicular from R R' will strike O N. Hence ft=OM + MR
and p 2 =0 M-M R and the direction of p l is M X, which bisects
< R M N. The principal stresses are shown in the figure.
If the principal stresses are found at various points on the
cross-section of a loaded beam, the results will be as shown in
Fig. 70, c. At the top and bottom simple stresses of compres-
sion and tension exist, while at the neutral axis there is a pull and
thrust at 45 to the axis, each equal to the unit shear on the ver-
tical or horizontal plane. A network of lines intersecting at right
angles, such as is shown in Fig. 72,
will give the direction of the princi-
pal stresses at any point of the beam
and may be called lines of principal
stress. Those convex upward give the direction of the compres-
sive forces. At the section of maximum bending moment, at
which section the shear is zero, the curves are horizontal and the
stress is greater than at any other point on the curve; from there
the stress diminishes to zero at the edge of the beam. The lines
of principal stress also show the traces of surfaces on which the
stress is always normal.
Fig. 72
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INTERNAL STRESS: CHANGE OF FORM.
Vol
169. Rankine's Theory of Earth Pressure. — Any embank-
ment or cutting keeps its figure both by the friction and by the
adhesion between the grains of earth. The latter is variable and
uncertain in amount and may even be entirely absent, for which
reason it is neglected in the present discussion. A bank of earth
is, therefore, to be conceived as a granular mass, retaining its
shape by friction alone, as is the case with a pile of dry sand.
When any such granular material is heaped up in a pile, the sides
slide until they assume a certain definite slope depending upon
the material. The greatest angle with the horizontal which the
sides can be made to take is the angle 0} repose, and its tangent
is the coefficient of friction of the material. -
It is evident that in a granular mass there is pressure on any
plane which can be passed through it, and froni § 161 it is seen
that, if the principal stresses in the mass are unequal, the stress
on any plane, not a principal plane, is oblique. But in earth, the
obliquity of the pressure on any plane cannot exceed the angle
of repose, <f>, or slidkg would take place along that plane. If
X
Fig. 73
Fig. 73 shows the plane upon which the direction of pressure is
most oblique (see §165), it is necessary for equilibrium that the
angle ROM shall not exceed <j>. TJien
sinROM=$^ 2 <sin£.
Pl+p2
By rearranging the terms, the ratio between the principal pres-
sures is found to be
P*>±
-sin <j>
pi i+sin<£ #
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l82 STRUCTURAL MECHANICS.
In the figure, pi is taken larger than p 2 \ if p 2 is larger than p i9
the triangle will lie on the other side of the normal and the ratio
becomes
P2 ^ 1 4- sin <f>
Pi — 1 —sin <f>'
If />i is the principal pressure acting vertically and due to the
weight of the earth itself, its conjugate pressure, p 2y must he
between the values given by the two preceding equations. Its
exact amount can be determined by the principal of least resist-
ance, for piy which is caused by the action of gravity, produces
a tendency for the earth to spread laterally, which tendency
gives rise to the pressure, p 2 . As p2 is caused by pi, p 2 will
not increase beyond the least amount sufficient to balance pu
hence
p2 _ 1 —sin <f>
pi i+sin^'
and in Fig. 73 the angle R O M is equal to <f>.
In the preceding discussion pi has been considered to act on
a horizontal, and p 2 on a vertical plane, and such will be the case
when the surface of the ground is level, but if the ground slopes,
the pressure on a plane parallel to the surface is vertical and the
direction of pi inclines slightly from the vertical, as shown in the
following example:
Example. — Find the pressure on the back of the retaining-wall
shown in Fig. 74, and also the resultant pressure on the joint C O'. The
pressure on the plane O O' passed parallel to the surface of the ground
is vertical and due to the weight of the earth upon it of depth K O.
But the prism of earth resting on one square foot of the plane has a
smaller horizontal section than one square foot, and the ratio of the
unit vertical pressure on the plane through O to the weight of a verti-
cal column of earth one square foot in cross-section will be that of the
normal O H to the vertical O K. Hence OR(=OH) represents in feet
of earth the pressure per square foot on the plane O O'. Draw a line,
O S, making the angle of repose, <f> } with the normal and by trial find
on O H a centre, M, from which may be drawn a semicircle tangent
to the line, O S, and passing through R. 4 Then O M=i(^i-f p 2 ) and
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INTERNAL STRESS: CHANGE OF FORM.
183
M R=£(#i~^2)> and M X gives the direction of pi, the unit of pressure
being the weight of one cubic foot of earth.
The principal stresses are now known and the pressure on the back
of the wall can be found. To find the pressure on O' B at O', draw
O' A parallel to M X and O' M' normal to the back of the wall. Lay
off (y M' equal to O M, and M' R' equal to M R and making the same
angle with O' A as O' M' does. Then R' O' gives the direction of pres-
sure on the back of the wall and, when measured by the scale of the
drawing and multiplied by the weight of a cubic foot of earth, gives
the pressure in pounds per square foot at O'.
As the pressure on the back increases regularly with the distance
below the surface of the ground, the centre of pressure will be at D,
one-third of the slant height from O', and the total earth pressure
against one foot in length of the wall will be P= *X O' B X O' R'X weight
of a cubic foot of earth. If W is the weight of one foot length of wall
applied at the centre of gravity, G, the resultant of P and W is the
resultant pressure on the bed-joint, O' C, and the point where it cuts
the joint should lie within the middle third.
170. Change of Volume. — If ^=unit stress per square inch
on the cross-section of a prism, and A is the resulting stretch or
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184 STRUCTURAL MECHANICS.
shortening per unit of length, then by definition E=p+X, if
P does not exceed the elastic limit.
When a prism is extended or compressed by a simple longi-
tudinal stress, it contracts or expands laterally, Fig. 75. This
contraction or expansion per unit of breadth may be written
TA-^m in which m, the ratio of longitudinal extension to lateral
contraction, is a constant for a given material, and for most
solids lies between 2 and 4.
A simple longitudinal tension p then accompanies a
Longitudinal stretch = X = p+E per unit of length, and a
Transverse contraction =-]-rw=-^ mE per unit of
breadth.
For ordinary solids A is so small that it makes no difference
whether it is measured per unit of original or per unit of stretched
length. The original length will be used here.
The new length of the prism is /(i+A) and the cross-section
is S(i— \+m) 2 . The volume has changed from SI to
* S/(i+A — 2\+m) nearly, if higher powers of A
*| than the first are dropped, since the unit deforma-
| tions are very small. The change of unit volume
t is therefore A( 1 j. Thus, if m is nearly 4,
I for metals, the change of volume of one cubic
dd-y) "*
I unit is JA nearly, the volume being increased
for longitudinal tension. If there were no change
of volume, m would be 2, as is the case for india
rubber for small deformations. Similarly, for compression the
change of the unit volume is nearly— \\ for metals, the volume
being diminished.
Example. — Steel, £=29,000,000; ^=20,000 lb. per sq. in. tension;
the extension will be of its initial length, the lateral contraction
1,450
will be about —z — of its initial width, and its increase of volume about
5,800
2,900
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INTERNAL STRESS: CHANGE OF FORM.
185
171. Effect of Two Principal Stresses. — Denote the stresses
by p\ and p2, treated as tensile. If they are compressive, reverse
the signs.
Under the action of pi there will be the following stretch
of the sides per unit of length : Fig. 76,
parallel to O C,
parallel to OB and O A, - -%.
Under the action of p 2 there will be
parallel to O B,
h
E'
/A 1
/
B
c
/
"J
Fig. 70
parallel to O A and O C, - ~.
nth,
Adding the parallel changes or stretches
parallel to O C, X x =4(/>i -— ) ;
parallel to O B, X 2 =Up 2 -^\ ;
parallel to O A, Aa= -~^(Pi + p2).
If pi and p 2 are equal unit stresses, but of opposite signs,
the changes of length become
E\ 1+ h) ; "E( I+ y ; andzer °;
or, putting either of these two changes equal to A, the lengths
to the sides of the cube originally unity per edge will be i+jl,
1 —X, and 1, and the volume, neglecting X 2 , is unchanged.
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1 86 STRUCTURAL MECHANICS.
172. Effect of Two Shears. — The cube shown in Fig. 77
has been deformed by the action of two equal and opposite prin-
cipal stresses, and a square, traced on one side of the original
cube, has been distorted to a rhombus, the angles of which are
greater and less than a right angle by the amount, <j>. By § 169,
two equal principal stresses of opposite sign are equivalent to
two unit shears of the same amount per square inch on
planes at 45 with the principal axes; hence the distortion of
the prism, whose face is the rhombus, results from two equal
shears at right angles.
Now one-half the angle £*— has for its tangent £(1 —A) -f-
%(i+X); hence
1 —^ ,,, ,n l —tan h<f> ,
—ptan *(**-# = i+tan — or A=tan tf.
But as ^ is small, A= \<j> } or <f> = 2X.
Therefore a stretch and an equal shortening, along a pair of
rectangular axes, are equivalent to a simple distortion relatively
to a pair of axes making angles
of 45 with the original axes ; and
the amount of distortion is double
that of either of the direct changes
of length which compose it. This
fact also appears from the consid-
eration that a distortion of a
square is equivalent to an elon-
gation of one diagonal and a
shortening of the other in equal proportions.
Example. — For steel, as before, A= , <b= — =4' 45", if *,=
1,450' ^ 725 * ^ ' tl
—p2= 20,000 lb.
173. Modulus of Shearing Elasticity. — Similarly, equal shear-
ing stresses q on two pairs of faces of a cube, in directions par-
allel to the third face, will distort that third face into a rhombus,
each angle being altered an amount <j>, there being distortion of
shape only, and not change of volume, Fig. 78.
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INTERNAL STRESS: CHANGE OF FORM.
187
Under the law which has been proved true within the elastic
limit, and the definition of the modulus of
elasticity, § 10, a modulus of transverse (or
shearing) elasticity, C, also called coefficient
of rigidity, as E may be called coefficient of
stiffness, may be written, C = q+<f>.
As these two unit shears are equivalent to
a unit pull and thrust of the same magnitude
per square inch, at right angles with each
other and at 45 with these shears, the case is identical with the
preceding one. Then
Fig. 78
= 2A, and X
-1K>
2p m-T-i
But, as p = q=C0,
<f> 2 m-T-i
For iron and steel m is nearly 4, which gives C=\E. For
wrought iron and steel, C is one or two one-hundredths less than
0.4E. Some use $£. C= 11,000,000 is a fair value.
174. Stress on One Plane the Cause of Other Stresses. — The
elongation produced by a pull, the shortening produced by a
thrust, and the distortion due to a shear can be laid off as graphi-
cal quantities and discussed as were unit stresses themselves. All
the deductions as to stresses have their counterparts in regard to
changes of form. There has been found an ellipse of stress for
forces in one plane, when two stresses are given. Also, when
three stresses not in one plane are given, there is an ellipsoid of
stress which includes all possible unit stresses that can act on
planes in different directions through any point in a body. So
there is an ellipse or ellipsoid that governs change of form.
Whether the movement of one particle towards, from or by
its neighbor sets up a resisting thrust, pull or shear, or the appli-
cation of a pressure, tension, or shear is considered to cause a
corresponding compression, extension, or distortion, the stresses
and the elastic change of form coexist. Hence it follows that,
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188 STRUCTURAL MECHANICS.
when a bar is extended under a pull and is diminished in lateral
dimensions, a compressive stress acting at right angles to the pull
must be aroused between the particles, and measured per unit of
area of longitudinal planes, together with shears on some inclined
.sections.
That such a state of things can exist may be seen from the
following suggestions. It may be conceived that the particles of
a body are not in absolute contact, but are in a
state of equilibrium from mutual actions on one
another. They resist with increasing stress all
attempts to make them approach or recede from
each other, and, if the elastic limit has not been
exceeded, they return to their normal positions
when the external forces cease to act. The par-
ticles in a body under no stress may then be con-
ceived to be equidistant from each other. The
smallest applied external force will probably cause change in their
positions.
If, in the bar to which tension is to be applied, a circle is
drawn about any point, experiment and what has been stated
about change of form in different directions show, that the diam-
eter in the direction of the pull will be lengthened when the force
is applied, the diameter at right angles will be shortened, and the
circle will become an ellipse. In Fig. 79, particle 1 moves to 1',
2 to 2', 4 to 4', and 7 to 7'. As they were all equidistant from
o in the beginning, 1 in moving to 1' offers a tensile resistance,
7 resists the tendency to approach o, while a particle near 4, mov-
ing to 4', does not change its distance from o, but moves laterally
setting up a shearing stress. A sphere will similarly become an
ellipsoid.
175. Actual Resulting Stresses. — Let — />i be the unit tension
in the direction 0-1, and + p\+m the accompanying unit thrust
in the direction 7-0. If a pull is applied to the solid in the direc-
tion 0-7, which develops —p2 tension in that direction on the
plane 0-1, and +p2+m thrust in the direction 1-0, the resultant
unit tension along 0-1 on the plane 0-7 will be —p\+p2+m, and
along 0-7 on the plane 0-1 will be —p2 + pi +fn. It follows that
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INTERNAL STRESS: CHANGE OF FORM. 189
the tension in the direction 0-1 will be less than when the first pull
was acting alone. Hence a plate is stronger to resist two pulls
at right angles than when subjected to one only. The opposite
deduction can be drawn if one principal stress is of opposite sign
to the other.
A boiler plate has a tension in a tangential direction, that is,
on a linear inch of longitudinal element, of pr, or of pr+t per
square inch, where p = steam pressure per square inch, r= radius
in inches, and / = thickness of plate. On a circumferential inch
the pull is one-half as much. Then, by § 170, and what has
been stated above, if w=4,
Pi=-pr p x +m=-\pr
p 2 =-\pr p2+m=~\pr
pi' = pi-p 2 +fn=-lpr
p 2 '=p2—pi+in=-\pr
Hence the true unit tension is less than the apparent tension by
12J per cent, and the boiler is stronger than it would be if the
longitudinal tension from the steam pressure on the heads did
not exist.
If tension is applied to the ends of a rectangular prism, and
external compre sion is added to all four sides, the true unit ten-
sion is much increased, or the piece is decidedly weaker in resist-
ing the pull.
Example. — At a certain point in a conical steel piston there
exist principal stresses of 3,160 and 1,570 lb., of opposite signs. Then
i>i'=3 J I 6o+i- I »57o=3»55o lb -J /^i^o+i^^Go^^oo lb.
Since test experiments to determine tensile and compressive
strength are made by the application of a single direct force, the
values so determined are compatible with the existence of the
opposite stress on planes at right angles with the cross-section.
Hence the working stresses for any material may fairly be con-
sidered to be a little higher than ordinary experiments show,
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190 STRUCTURAL MECHANICS.
provided account is taken at the same time of all the stresses
which act on a particle.
176. Cooper's Lines. — Steel plate as it comes from the mill
has a firmly adhering but very brittle film of oxide of iron on the
surface. This film is dislodged by the extension of a test speci-
men in tension when the yield-point is passed. If a hole is
punched at moderate speed in a steel plate, so that the particles
under the punch have some opportunity to flow laterally under
the compression, there will be a radial compressive stress in all
directions outwardly from the circumference of the hole, and a
tensile stress circumferentially. These opposite principal stresses
cause shearing planes to exist whose obliquity depends upon the
relative magnitudes of the principal stresses by § 163. The
scale breaks on these lines of shear and there result curves where
the bright metal shows through, branching out from the hole,
intersecting and fading away. The process of shearing a bar
will develop the same curves from the flow of the metal on the
face at the cut end. They are known as Cooper's lines. These
lines show that deformation takes place at considerable distances
from the immediate point of shearing or punching.
Examples. — 1. A pull of 1,000 lb. per sq. in. and a thrust of 3,000
lb. per sq. in. are principal stresses. Find the kind, direction, and
magnitude of the stress on a plane at 45 with either principal plane.
2. Find the stress per running unit of length of joint for a spiral
riveted pipe when the line of rivets makes an angle of 45 with the
axis of the pipe and when it makes an angle of 6o°. 0.707 p; 0.5 p.
3. A rivet is under the action of a shearing stress of 8,000 lb. per
sq. in. and a tensile stress, due to the contraction of the rivet in cooling,
of 6,000 lb. per sq. in. Find pi and p2.
/>!=- 11,540 lb.; /> 2 = + 5>54olb.
4. A connecting plate to which several members are attached has
a unit tension on a certain section of 6,500 lb. at an angle of 30 with
the normal. On a plane at 6o° with the first plane the unit stress of
5,000 lb. compression is found at 45 with its plane. Find the principal
unit stresses and the shear. —6,600; +4,800; 5,700.
5. Assuming the weight of earth to be 105 lb. per cu. ft. and the
horizontal pressure to be one-third the vertical, what is the direction
and unit pressure per square foot on a plane making an angle of 15
with the vertical at a point 12 ft. under ground, if the surface is level?
515 lb.; 39$° with the horizon.
6. A stand-pipe, 25 ft. diam., 100 ft. high. The tension in lowest
ring, if | in. thick, is 7,440 lb. per sq. in. If plates range regularly
from I in. thick at base to } in. at top, neglecting lap, the compression
at base will be about 215 lb. per sq. in. For a wind pressure of 40
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INTERNAL STRESS: CHANGE OF FORM. 191
lb. per sq. ft., reduced 50% for cylindrical surface, and treated as if
acting on a vertical section, M at base= 2,500,000 ft.-lb. Compression
on leeward side at base=485 lb. per sq. in. If ^=—7,440 lb., p2= s
215+485=700 lb., find the stress and its inclination for a plane at 30
to the vertical? —6,193 lb.; 34 40'.
Prove that the shearing plane is 17 04' from horizontal, and that
the shear is 2,284 lb- per sq* hi*
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CHAPTER XII.
RIVETS: PINS.
177. Riveted Joints. — There are four different ways in which
riveted joints and connections may fail. The rivets may shear
off; the hole may elongate and the plate cripple in the line of
stress; the plate may tear along a series of rivet-holes, more or
less at right angles to the line of stress; or the metal may fracture
between the rivet-hole and the edge of the plate in the line of
stress. From the consideration that a perfect joint is one offering
equal resistance to each of these modes of failure, the proper pro-
portions for the various riveted connections are deduced.
178. Resistance to Shear. — The safe resistance of a rivet to
shearing off depends upon the safe unit shear and the area of the
rivet cross-section, which varies as the square of the diameter of
the rivet. When one plate is drawn out from between two others,
a rivet is sheared at two cross-sections at once, and is twice as
effective in resisting any such action. Rivets so circumstanced
are said to be in double shear, and their number is determined
on that basis.
179. Bearing Resistance. — The resistance against elongation
of the hole or crippling the plate depends on the safe unit com-
pression and what is known as the bearing area, the thickness of
the plate multiplied by the semicircumference of the hole. As
the semicircumference varies as the diameter, it is more con-
venient, and sufficiently accurate, to use the product of the thick-
ness of the plate and the diameter of the rivet with a value of
allowable unit compression about fifty per cent greater than
usual. In practice the bearing values is always given in terms of
the diameter.
180. Resistance of Plate. — The resistance to tearing across
the plate through a line of holes, or in a zigzag through two lines
192
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RIVETS: PINS. 193
of holes in the same approximate direction, depends on the safe
unit tensile stress multiplied by the cross-section of the plate
after deducting the holes. If the transverse pitch, or distance
between centres of rivets, is considerable, an assumption of
uniform distribution of tension on that cross-section is not likely
to be true.
The resistance of the metal between the rivet-hole and the
edge of the plate in the line of stress is usually taken as the safe
unit shear for the plate multiplied by the thickness and twice
the distance from the rivet- hole to the edge. Some, however,
consider that the resisting moment of the strip of metal in front
of the rivet-holes is called into action.
181. Bending: Friction. — There are those who advise the com-
puting of a rivet shank as if it were subjected to a bending moment.
If the rivet fills the hole and is well driven, there is no bending
moment exerted on it, unless it passes through several plates.
As practical tests have shown that rivets cannot surely be made
to fill the holes, if the combined thickness of the plates exceeds
five diameters of the rivet, this limitation will diminish the im-
portance of the question of bending.
No account is taken of the friction induced in the joint by
its compression and the cooling of the rivet, and such friction
gives added strength. As the rivet is closed up hot, the shank
is under more or less tension when cold. Moreover, the head
is not given the thickness required in the head of a bolt under
tension, and therefore rivets are not available for any more ten-
don, and should not be used for that purpose. Tight-fitting
turned bolts are required in such a case.
182. Spacing. — The rivets should be well placed in a joint
or connection, in order to insure a nearly uniform distribution of
stress in the piece; they should be symmetrically arranged, be
placed where they can be conveniently driven, and be spaced
so that the holes can be definitely and easily located in laying
out the work.
183. Minimum Diameter of Rivets. — The punch must have
a little clearance in the die. The wad of metal shears out below
the punch with more ease and with less effect on the surrounding
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1 94 STRUCTURAL MECHANICS,
metal when it can flow, as it were, a little laterally, and it then
comes out as a smooth frustum of a cone with hollowed sides,
reminding one of the vena conlracta. The punch must also be a
little larger than the rivet, to permit the ready entrance of the
rivet-shank at a high heat. The diameter of the hole is com-
monly computed at J inch in excess of the nominal diameter
of the rivet; but the rivet is treated as if of its nominal diameter.
One other consideration has weight in determining the minimum
diameter of the rivet. If the diameter of the rivet is less than
the thickness of the plate, the punch will not be likely to endure
the work of punching. A diameter one and a half times the
thickness of the plate is often thought desirable.
184. Number and Size of Rivets. — Formulas are of little or
no value in designing ordinary joints and connections. Boiler
joints and similar work can be computed by formulas, but to no
great advantage. Tables are used which give what is termed
the shearing value of different rivet cross-sections in pounds
for a certain allowable unit shear, and the bearing or compression
value of different thicknesses of plate and diameters of rivet for
a certain allowable unit compression. For a given thickness
of plate, that diameter of rivet is the best whose two values, as
above, most nearly agree. The quotient of the force to be trans-
mitted through the connection or through a running foot of a
boiler joint, divided by the less of the two practicable values, will
give the minimum number of rivets. Their distribution is governed
by the considerations previously referred to. Whether a joint
in a boiler requires one, two, or three rows of rivets depends upon
the number needed per foot.
Example. — Two tension bars, 6 in. by \ in., carrying 30,000 lb.,
are to be connected by a short plate on each side. Let unit shear be
7,500 lb. per sq. in., unit bearing 15,000 lb. on the diameter, and unit
tension 12,000 lb. The bearing value of a J-in. rivet in a £-in. plate
is 5,625 lb., its shearing value in double shear is 2X3,310=6,620 lb.
Hence 30,000^-5,625 = 6 rivets necessary. If these rivets can be so
arranged that a deduction of but one rivet-hole is necessary from the
cross-section of the tie, (6— (f-f J))£=2.56 sq. in. net section, which
will carry 30,750 lb. at 12,000 lb. unit tension. Each cover-plate can-
not be less than \ in. thick, and, as will be seen presently, should be
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RIVETS: PINS.
195
made a little more. The length will depend on the distribution of the
rivets, to be taken up next.
185. Arrangement of Rivets. — Long joints, under tension,
like those of boilers, are connected by one or more rows of rivets,
I
000
000
000
OO
L&J
GF1
o 1; o o
o o I o o
olio o
[
)
ojjo o\
000 II H 0]
000 1 ol
OOOO oib 0/
11 r
Fig. 80
as shown at A and B, Fig. 80. If more than one row is needed,
the rivets are staggered, as at B, and the rows should be separated
such a distance that fracture by tension is no more likely on a
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*9* STRUCTURAL MECHANICS.
zigzag line than across a row. Experiments have shown that a
plate will break along a zigzag line such as is shown at B unless
the net length of that line exceeds by about one-third the net
length of a line through a row. To prevent tearing out at the
edge of the plate, the usual specification of at least one and a
half rivet diameters from centre of hole to edge of plate will
s;uffice.
The tendency of a lap-joint to cause an uneven distribution
of stress by reason of bending, and the same tendency when a
single cover is used, is shown at C. The increase of stress thus
caused should be offset by increased thickness of plate. A cover
strip on each side is preferable if not objectionable for other
reasons.
In splicing ties, D shows a bad arrangement, the upper plan
failing to distribute the stress evenly across the tie, and the lower
plan wasting the section by excessive cutting away. The rivets
at E are well distributed across the breadth, and weaken the tie
by but one hole, as only two-thirds of the stress passes the section
reduced by two holes; and, unless the net section at this place
is less than two-thirds of the section reduced by one hole, it is
equally strong. Thus b-d=$(b — 2</), or a breadth equal to
or greater than four diameters will satisfy this requirement.
The covers, however, will be weakened by two holes, and hence
their combined thickness, when two are used, should exceed the
thickness of the tie.
F similarly is better than G, and the tie at F is again weakened
by but one hole. The sectional area of the plate shown at H
is diminished by two holes at m and four at n, but the stress
on section n is less than that on m by the stress the rivets at m
transmit to the splice plate. Consequently, in designing a splice,
if the area cut out by the two extra rivets at n multiplied by the
working stress in the plate does not exceed the working value
of the two rivets at m, the plate will be weakened by two holes
only. In the splice shown the sections at m and n will be equally
strong when b — 2<Z =-[£(&— ^d) or when b = iM.
As it is desirable to transmit all but the proper fraction of the
tension past the first rivet, the corners of the cover F or H are
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RIVETS: PINS. 197
clipped off, thus increasing the unit tension in the reduced section
and increasing its stretch to correspond more nearly with the
unit tension and elongation of the tie beneath. The appearance
of the connection is also improved.
It is not desirable to make splice plates as short as possible,
because a short splice is likely to be weak. In splicing members
built up of shapes this is especially the case, as a uniform dis-
tribution of stress over the whole cross-section is not easily secured
when the stress passes into the member within a length less
than its width. Short splice plates may be lengthened without
increasing the number of rivets by using a greater pitch.
186. Remarks. — If the member is in compression, the holes
are not deducted, since the rivets completely fill the holes; and
the strength is computed on the gross section. Unless special
care is exercised in bringing two connected compression pieces
into close contact at their ends, good practice requires the use of
a sufficient number of rivets at the connection to transmit the
given force.
Rivet-heads in boiler work are flat cones. In bridge and
structural work they are segments of spheres, known as button
heads, and are finished neatly by means of a die. These heads
may be flattened when room is wanting, and countersunk heads
are used where it is necessary to have a finished flat surface.
Members of a truss which meet at an angle are connected
by plates and rivets. The axes of the several members should
if possible intersect in a common point. If they do not, moments
are introduced which give rise to what are known as secondary
stresses, as distinguished from the primary stresses due to the
direct forces in the pieces of the frame. Such secondary stresses
may be of considerable magnitude in an ill-designed joint.
It is desirable to arrange the rivets in rows which can be
easily laid out in the shop, and to make the spacing regular,
avoiding the use of awkward fractions as much as possible.
Commercial rivet diameters vary by eighths of an inch, £-,
}-, and |-in. rivets being the ones frequently used. As much
uniformity as possible in the size of rivets will tend tc economy
in cost.
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198 STRUCTURAL MECHANICS.
187. Structural Riveting.— The following rules for structural
work are in harmony with good practice:
Holes in steel f in. thick or less may be punched; when
steel of greater thickness is used, the holes shall be subpunched
and reamed or drilled from the solid.
The diameter of the die shall not exceed that of the punch
by more than A of an inch, and all rivet-holes shall be so accu-
rately spaced and punched that, when the several parts are
assembled together, a rivet A in. less in diameter than the hole
can generally be entered hot into any hole.
The pitch of rivets, in the direction of the stress, shall never
exceed 6 in., nor 16 times the thickness of the thinnest plate
connected, and not more than 30 times that thickness at right
angles to the stress.
At the ends of built compression members the pitch shall not
exceed 4 diameters of the rivet for a length equal to twice the
width of the member.
The distance from the edge of any piece to the centre of a
rivet-hole must not be less than ij times the diameter of the
rivet, nor exceed 8 times the thickness of the plate; and the
distance between centres of rivet-holes shall not be less than 3
diameters of the rivet.
The effective diameter of a driven rivet will be assumed to be
the same as its diameter before driving; but the rivet- hole will
be assumed to be one-eighth inch diameter greater than the un-
driven rivet.
In structural riveting these relationships between unit working
stresses are very commonly used:
Bearing stress = i£ X tensile stress ;
Shearing stress = \ X bearing stress = f X tensile stress.
The shearing area of the rivets, therefore, should exceed by
one-third the net area of the tension member they connect. See
§§ 146, 147, 148.
188. Boiler-riveting. — Boiler work admits of standardization
much more readily than structural work and standard boiler
joints, which make the tensile strength of the net plate equal to
the strength of the rivets, have been very generally adopted. A
triple-riveted boiler joint is shown in Fig. 80, L. The most
notable point of difference between boiler and structural riveting
is that it Is not customary to consider the bearing of rivets in boiler
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RIVETS: PINS. 199
work. Rivets are figured for shear only and in double shear are
considered to have but one and three-quarters time the value of
rivets in single shear, instead of twice the value, as in structural
work. Custom in this country makes the allowable unit shear
on rivets approximately two-thirds the unit tensile stress in the
plate, but the British Board of Trade rule gives about four-fifths.
The diameter of the rivets used should be about twice the thick-
ness of the plate. In ordinary cases there is no danger that the
rivets will be too far apart to render the joint water- or steam-
tight, when the edge of the plate is properly closed down with a
calking-tool.
189. Pins: Reinforcing Plates. — The pieces of a frame are
frequently connected by pins instead of rivets. The axes of the
several pieces are thus made to meet in a common point, if the
pin-hole is central in each member. Pins are subjected to com-
pression on their cylindrical surfaces, to shear on the cross-section,
and to bending moments. The compression on the pin-hole is
reduced to the proper unit stress, if necessary, by riveting re-
inforcing plates to the sides of the members, as shown at K,
Fig. 80. A sufficient number of rivets to transmit the proper
proportion of the force must be used, with a due consideration
of the shearing value of a rivet and its bearing value in the re-
inforcing plate or the member itself, whichever gives the less value.
No more rivets should be considered as efficient behind the pin
than the section of the reinforcing plate each side of the pin-hole
will be equivalent to.
When the pin passes through the web of a large built member,
such as a post or a top chord of a bridge, the web is often so thin
that more than one reinforcing plate on either side is needed.
It is then economical to make the several plates of increasing
length, the shortest on the outside, and determine the number
of rivets in each portion accordingly.
Pin-plates should be made long for the same reason as given
for making long splice plates. The longest plate is sometimes
required to extend 6 in. inside the tie-plates so that the stress
may be transferred to the flanges and not overtax the web. See
K, Fig. 80.
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200
STRUCTURAL MECHANICS.
190. Shear and Bearing. — The shear at any section of the pin
is found from the given forces in the pieces connected. The
resultant of the forces in the pieces on one side of any pin section
will be the shear at that section. As the pin will probably not
fit the hole tightly (a difference of diameter of one-fiftieth of an
inch being usually permitted), the maximum unit shear will be
four-thirds of the mean (§ 72). Specifications frequently give a
reduced value for mean unit shear, which provides for this un-
equal distribution. Bearing area is figured as if projected on
the diameter.
191. Bending Moments on Pins. — At a joint where several
pieces are assembled, the resisting moment required to balance
the maximum bending moment on the pin caused by the forces
in those pieces will generally determine the diameter of the pin.
In computing the bending moments, the centre line of each piece
or bearing is considered the point of application of the force
which it carries. This assumption is likely to give a result some-
what in excess of the truth, as any yielding tends to diminish
the arm of each force.
The process of finding the bending moments will be made
clear by an illustration. Fig. 81 shows the plan and elevation
A
IV
wooo
i s
B
D
4-
V
)0
t^e=^=™=
B
l
V <
A
-,_
Fig. 81
of the pieces on a pin, with the forces and directions marked.
The thickness of the pieces, which are supposed to be in contact,
is also shown. The joint must be symmetrically arranged, to
avoid torsion, and simultaneous forces must be used, which reduce
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RIVETS: PINS. 201
to zero for equilibrium. As the joint is symmetrical, the com-
putation is carried no farther than the piece adjoining the middle.
Resolve the given forces on two convenient rectangular axes,
here horizontal and vertical. Set the horizontal components ia
order in the column marked H, the vertical ones in the column
marked V. Their addition in succession gives the shears, marked
F. The next column shows the distance from centre to centre
of each piece. Fdx is then the increment of bending moment;
and the summation of increments gives, in the column M, the
bending moment at the middle of each piece, from the horizontal
and from the vertical components respectively. The square root
of the sum of the squares of any pair of component bending
moments will be the resultant bending moment at that section.
It is comparatively easy to pick out the pair of components which
will give a maximum bending moment on the pin. Equate this
value with the resisting moment of a circular section and find
the necessary diameter.
H.
F.
dx.
Fdx.
M.
A
+ 10,000
+ 10,000
1*
B
—40,000
— 30,000
i
+ 11,250
+ 11,250
C
—30,000
«
-22,500
-",250
D
+ 15,000
-15,000
i
-18,750
—30,000
E
A
+ 15,000
V.
- 5*780
- 5,78o
1*
-I3.I25
-43,125
B
- 5>78o
i
- 6,503
- 6,503
C
- 2,890
- 8,670
f
~ 4,335
-10,838
D
- 8,670
i
~ 5.419
-16,257
E
+ 8,670
" 7,586
-23,843
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aoa STRUCTURAL MECHANICS.
M at D=V(3o,ooo 2 + i6,2S7 2 ) ; M at E=x/(43>i25 2 + 23>843 2 )-
The latter is plainly the larger, and is 49,210 in.-lb.
The pieces can be rearranged on this pin to give a smaller
moment. The maximum moment is not always found at the
middle.
The bending moment at any point of the beam or shaft, when
the forces do not lie in one plane, can be found in the same way.
A solution of the above problem by graphics can t>e found
in the author's Graphics, Part II, Bridge Trusses.
Examples. — 1. A tie-bar J in. thick and carrying 24,000 lb. is
spliced with a butt-joint and two covers. If unit shear is 7,500 lb. #
unit bearing on diameter is 15,000 lb., and unit tension is 10,000 lb.,
find the number, pitch, and arrangement of J-in. rivets needed, and
the width of the bar.
2. The longitudinal lap-joint of a boiler must resist 52,000 lb. ten-
sion per linear foot. If the unit working stress for the shell is 12,000
lb. and the other stresses as above, what size of rivet is best, for triple
riveting, what the pitch, and the thickness of the shell ?
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CHAPTER Xm.
ENVELOPES.
192. Stress in a Thin Cylinder. — Boilers, tanks, and pipes
under uniform internal normal pressure of p per square inch.
Conceive a thin cylinder, of radius r, to be cut by any diame-
tral plane, such as the one represented in Fig. 82, and consider
the equilibrium of the half cylinder, which is illustrated on the
left. It is evident that, for unity of distance along the cylinder,
the total pressure on the diameter, 2pr, must balance the sum
of the components of the pressure on the semi-circumference in
a direction perpendicular to the diameter. This pressure, 2pr,
uniformly distributed over the diameter, must cause a tension
T in the material at each end to hold the diameter in place. Hence
T-fr.
As all points of the circle are similarly situated, the tension in
the ring at all points is constant and equal to pr. If the thick-
ness is multiplied by the safe working tension / per square inch,
it may be equated with pr, giving
Required net thickness ^pr+f.
In a boiler or similar cylinder made up of plates an increase
of thickness will be required to compensate for the rivet-holes.
II a is the pitch, or distance from centre to centre, of consecu-
203
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one
204 STRUCTURAL MECHANICS.
tive rivets in one row along a joint, and d the diameter of the
rivet-hole, the effective length a to carry the tension is reduced
to a—d, and the gross thickness of plate must not be less than
pr a
} a—d'
Example. — The circumferential tension in a boiler, 4 ft. diameter,
carrying 120 lb. steam pressure is 120-24=2,880 lb. per linear inch of
length of shell, which will require a plate — in. thick (net), if / is
not to exceed 10,000 lb. per sq. in. Net thickness =^ in. If a longi-
tudinal joint has J-in. rivet-holes, at 2J in. pitch, in two rows, the
thickness of plate must not be less than — ! t^tV hi.
r 10,000-1$ 10
193. Another Proof of the value of T may be obtained as
follows: The small force on arc ds=*pds. The vertical compo-
nent of this force =pds sin = pdx. The entire component on
pdx=2pr, which must be resisted
by the tension in the material at the two ends of the diameter.
The same result will be obtained graphically by laying off a
load ]ine = I pds, which become a regular polygon of an infinite
number of sides, i.e., a circle, with the lines to the pole making
the radii of the length pr.
The cylinder, under these circumstances, is in stable equi-
librium. If not perfectly circular, it tends to become so, small
bending moments arising where deviation from the circle exists.
Hence a lap-joint in the boiler shell causes a stress from the
resisting moment to be combined with the tension at the joint.
The above investigation applies only to cylinders so thin that
the tension may be considered as distributed uniformly over the
sect on of the plate.
For riveting see Chapter XII.
194. Stress in a Right Section. — The total pressure from p
on a right section of the cylinder is 7& 2 p } which will also be the
resultant pressure on the head in the direction of the axis of the
cylinder, whether the head is flat or not. This pressure causes
tension in every longitudinal element of the cylinder, or in every
cross-section. As this cross-section is 2izr Xthickness, the longi-
tudinal tension per linear inch of a circumferential joint is
x
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ENVELOPES. 205
7cr 2 p +27& =bpr, or one-half the amount per linear inch of a
longitudinal joint. Hence a boiler is twice as strong against
rupture on a circumferential joint as on a longitudinal joint,
and hence the longitudinal seams are often double riveted while
the circumferential ones are single riveted.
195. Stress in any Curved Ring tinder Normal Pressure. —
The stress in a circular ring of radius r, under internal or ex-
ternal normal unit pressure, p, is pr per linear unit of section of
the ring, being tension in the first case and compression in the
second. Similarly, in single curved envelopes in equilibrium
under normal pressure (that is, envelopes in which the stress acts
in the direction of the shell) the stress at any point per linear
unit along an element is equal to pp, in which p is the radius
of curvature of the cu ve cut out by a plane normal to the element
and passing through the given
point. Fig. 83 shows the trace
of an arc of shell of width unity
and of length pdO acted upon from
within by a normal pressure, p.
Then P=*ppdQ and for equilibrium
the sum of the components acting
in the direction of P must be zero
or P = 2T sin \dO=TdO. Hence T=pp.
196. Thin Spherical Shell : Segmental Head. — If a thin hollow
sphere of radius r* has a uniform normal unit pressure p applied
to it within, the total interior pressure on a meridian plane will
be n^p, and the tension per linear inch of shell will be
7^ 2 p^27zr , = \pr f .
If p is applied externally, the stress in the material will be com-
pression. It may be noted that the double curvature of the
sphere is associated with half the stress which is found in the
cylinder of single curvature having the same radius.
If a segment of a sphere is used to close or cap the end of
a cylinder or boiler, the same value will hold good. In this case
the radius f is greater than r for the cylinder.
If the segmental end is fastened to the cylinder by a bolted
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2o6 STRUCTURAL MECHANICS.
flange, the combined tension on the bolts will be icr*p, as this
is the total force on a right section of the 'cylinder.
The flange itself will be in compression. The pressure p
from below, in Fig. 84, causes a pull per circumferential unit
in the direction of a tangent at B, which pull has just been shown
to be equal to \pr*. It may be resolved into vertical and hori-
zontal components. The vertical component B C is, by § 194,
\pr. The horizontal component h must be proportioned to the
vertical component as A O to A B, the sides of the right-angled
^^_^ triangle to which they are respec-
~^^^ ^TT/^Nw tively perpendicular. As AO =
J_._._.a. 3* V(S*-r*),
\ i or h=\pV {'*-*).
s 1 As A is a uniform normal pressure
Fig. 84 i- 1 r -i /
applied from without (or tension
applied from within) in the plane of the flange, the compression
on the cross-section of the latter will be hr or £/*V(r' 2 -r 2 ), to
be divided by that cross-section for finding the unit compression.
Segmental bottoms of cylinders are sometimes turned inward.
The principles are the same.
Example. — A segmental spherical top to a cylinder of 24 in. diam-
eter, under 100 lb. steam pressure, has a radius of 15 in. with a versed
sine of 6 in. The tension in top=£- 100-15=750 lb. per linear inch.
If its thickness is J in., the stress per sq. in. is 3,000 lb. The total pull
on the flange bolts is 100-144- 22-5-7 = 45,260 lb. A J-in. bolt has
about 0.3 sq. in. section at bottom of thread, giving a tension value of
about 3,000 lb. if /= 10,000 lb. There would be needed some 15 bolts,
about 5^ in. centre to centre on a circumference of 26 in. diameter.
The compression in the flange is £• 100-12-9=5,400 lb. A 2Xj-in.
flange with a J-in. hole has a section £-ii=f sq. in,, giving a unit
compression in the flange of 1-5,400=8,600 lb. per sq. in.
A similar compression acts in the connecting circle between a
water-tank and the conical or spherical bottom sometimes built. See
§§ 204, 205.
197. Thick Hollow Cylinder. — If the walls of a hollow cylin-
der or sphere are comparatively thick, it will not be sufficiently
accurate to assume that the stress in any section is uniformly
distributed throughout it. If the material were perfectly rigid,
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ENVELOPES.
307
the internal or external pressure would be resisted by the imme-
diate layer against which the pressure was exerted, and the re-
mainder of the material would be useless. As, however, the
substance of which the wall is composed yields under the force
applied, the pressure is transmitted from particle to particle,
decreasing as it is transmitted, since each layer resists or neu-
tralizes a portion of the normal pressure and undergoes ex-
tension or compression in so doing.
198. Greater Pressure on Inside. — Let Fig. 85 represent the
right section of a thick hollow cylinder, such as that of a hydraulic
press. Let f\ and f2 be the internal and external radii in inches;
pi and p2 the internal and external normal unit pressures in
pounds per square inch, p\ being the greater; and p the unit
normal pressure on any ring whose radius is r.
If a hoop is shrunk on to the cylinder, p* will be the unit
normal pressure thus applied to the exterior of the cylinder.
The unit tensile stress found in a thin layer of radius r and
thickness dr will be denoted by /, and will be due to that portion
of p which is resisted by the layer and not transmitted to the next
exterior layer. The total tension on the radial section of a ring
lying between r x and r is piri—pr, since the pressure, p u on
the inside sets up a tension of ptf\ in the ring and the pressure,
p, acting on the outside of the ring sets up a compression of pr.
This total tension may also be expressed by
L
tdr. As p and r are variables, there is
obtained by differentiating the equation
pin -pr= I tdr,
or
-d(pr) =tdr,
pdr+rdp+ldr=o.
(1)
Another equation can be deduced from the
enlargement of the cylinder. The fibres or
layers between the limits r\ and r, being com-
pressed, will be diminished in thickness. The
compression of a piece an inch in thickness by a unit stress p
rig. 88
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208 STRUCTURAL MECHANICS.
will be p+E, § 10, and of one dr thick will be pdr+E. The
total diminution of thickness between r i and r, from what it was
i fr
at first, will therefore be -^ / pd\
r.
But the annular fibre or ring whose radius is r and length
2;rr has been elongated t+E per inch of length. Its length will
now be 2t& 1 1 + « ) and its radius r 1 1 +t=J- The internal radius
must similarly have become rA i + p J, where / is the value of /
for radius t\. The thickness f—f\ has now become r[*+p)
— ri[ 1 + ^;), and, by subtracting this value from r— fi, there is
found the diminution of thickness, ri~p— rjj- This expression
may be equated with the previous one for decrease of thickness,
or
r i- r i = if t r ^ r '
Since the first term is constant, there is now obtained by
differentiating this equation,
—d(tr)=pdr f or ldr+rdt+pdr=o. ... (2)
Add (1) and (2), and multiply by r to make a complete differ-
ential. Then integrate
2(t+p)rdr+r 2 (dt+dp) =0;
r 2 (/+^)=constant; .-. =r 1 2 (/+^ 1 )=r 2 2(/'+^ 2 ). . (3)
Again, subtract (1) from (2), and then integrate
dt—dp=o. /— ^=constant; .'. =}—pi=f—p2* • (4)
From (3) and (4) are obtained, by addition and subtraction,
, }-pir x * f + p K f-pifi 2 f+Pi ,.
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ENVELOPES. 209
If the internal radius is given, the external radius, and hence
the required thickness, f2— ri, is found by eliminating f from
(3) and (4),
*~4=££s) «>
If p2 is atmospheric pressure, it may be neglected when P\
is large. In that case
^ or „-„-,(^ : -.).
r 2 =fi
As f2 becomes infinite when the denominator of (6) is zero,
it appears that no thickness will suffice to bring / within the
safe unit stress, if p\ exceeds }+2p2.
These formulas do not apply to bursting pressures, nor to
those which bring / above the elastic limit; for E will not then
be constant. They serve for designing or testing safe construc-
tion.
Examples. — Cylinders of the hydraulic jacks, for forcing forward
the shield used in constructing the Port Huron tunnel, were of cast
steel, 12 in. outside diameter, 8 in. diameter of piston, with J in. clear-
ance around same; pressure 2,000 lb. per sq. in.
r£ 3 6#l6 / + 2,000
^""^"F^S? /=6 '° 3 °-
A cast-iron water-pipe at the Comstock mine was 6 in. bore,
2 J in. thick, and was under a water pressure of 1,500 lb. on the sq. in.,
or about 3,400 ft. of water. Here 7=2,770 lb. per sq. in. for static
pressure, while the formula for a thin cylinder gives 1,800 lb.
199. Greater Pressure on Outside. — In this case the direc-
tion or sign of / will be reversed, it being compression in place
of tension. From the preceding equations, without independent
analysis, by making / negative, there result:
-d(pr) - -tdr\ d(tr) =pdr.
pdr+rdp—tdr=o; tdr+rdt—pdr=o.
f*(p-t) -**&-[)-&&-?)-
t+P=f+pi=J' + p2.
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2IO STRUCTURAL MECHANICS.
The outer radius and pressure will now be taken as given
quantities, and the unit compression in the ring at any point
will be
*-^-=m <»>
which becomes, if P\ is neglected as small,
The external pressure p2 must be less than £(/+£i)> ^ f i * s
to have any value. It will be seen from / in (7) that the com-
pression is greatest at the interior.
Example, — An iron cylinder 3 ft. internal diameter resists 1,150
lb. per sq. in. external pressure. The required thickness, if /= 9,000
lb., is given by
*-^('-JS^^
uooo)
f2=2o.9 in. »Thickness=3 in.
200. Action of Hoops. — To counteract in a greater or less
degree the unequal distribution of the tension in thick hollow
cylinders for withstanding great internal pressures, hoops are
shrunk on to the cylinders, sometimes one on another, so that
before the internal pressure is applied, the internal cylinder is
in a state of circumferential compression, and the exterior hoop
in a state of tension. If the internal pressure on the hoop is
computed, for a given value of / in the hoop, and this pressure
is then used for p2 on the cylinder, the allowable internal pressure
Pi on the cylinder consistent with a permissible / in this cylinder
can be found. There is, however, an uncertainty as to the
pressure P2 exerted by the hoop.
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ENVELOPES. 211
Examples. — A hoop i in. thick is shrunk on a cylinder of 6 in.
external radius and 3 in. internal radius, so that the maximum unit
tension in the hoop is 10,000 lb. per sq. in. This stress, by § 198,
will be due to an internal pressure on the hoop of 1,530 lb. per sq. in.
For 7=6 J(I2*5±£) or <&-
\\io,ooo-/>i/ 3 6
10,000 -f^i
10,000— p{
This external pressure p2 on the cylinder will cause a compressive unit
stress in the interior circumference of the cylinder when empty, after
the hoop is shrunk on, of 4,080 lb., and will permit an internal pressure
in the bore of 8,448 lbs. per sq. in., consistent with /= 10,000 lb. For
— = , - . The cylinder alone, without the hoop, would
9 10,000—^1+3,000
allow a value of p\ given by — = — ! ^, or £1 = 6,000 lb. If the
r ° 9 10,000— pi
cylinder had been 4 in. thick, the internal pressure might have been
6,900 lb. The gain with the hoop, for the same quantity of material,
is 1,548 lb., or some 22%.
Hydraulic cylinder for a canal lift at La Louvtere, Belgium, 6 ft.
9 in. interior diameter, 4 in. thick, of cast iron, hooped with steel.
Hoops 2 in. thick and continuous. When tested, before hooping, one
burst with an internal pressure of 2,175 lb- V& S( l- m -» one at 2,280 lb.,
and a third at 2,190 lb. These results, if the formula is supposed to
apply at rupture, give an average tensile strength of 23,400 lb. per sq.
in. The hoops were supposed to have such shrinkage that an internal
pressure of 540 lb. per sq. in. would give a tension on the cast iron of
1,400 lb., and on the steel of 10,600 lb. per sq. in. The ram is 6 ft.
6} in. diam. and 3 in. thick, of cast iron, an example of the greater
pressure outside.
201. Thick Hollow Sphere. — Greater pressure on inside. Let
Fig. 85 represent a meridian section of the sphere. Suppose /, /,
etc., to be perpendicular to the plane of the paper. The entire
normal pressure on the circle of radius fi will be pirti 2 , and the
tension on the ring between radii fi and r will be ^(pir^—pr 2 ).
Any ring of radius r and thickness dr will carry 2itrtdr, and hence
is derived the first equation
n{P\r\ 2 - pf 2 ) =27: / rt dr, or —d{pr 2 ) « 2H dr.
/. r 2 dp+2pr dr+2rt dr=o.
The second equation will be the same as obtained for the cylinder.
-d(tr)=pdr, or rdt+tdr+pdr=o.
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212 STRUCTURAL MECHANICS.
Strike out the common factor r from the first equation, multiply
the second by 2, and subtract.
2rdt—rdp=o 9 or 2d/ — dp=*o.
•\ 2/—^= constant; .\ =2/— ^1 = 2/' — ^2- • • (9)
Again, add the first to the second and multiply by r 2 .
r*(dp+dt) + 3 r 2 dr(p+t) =0.
.\ r 3 (^+/)=constant; /. =r 1 3 (/ + /> 1 ) =r 2 3 (/ / +^ 2 ). . (10)
From (9) and (10),
. £/-^i ly* / + ft 2/-/^ fi 3 / + />i
3^3 3 ^3
-4^m <■»
These formulas are not applicable to bursting pressures for
the reason given before. For a finite value of r 2 , pi must be less
than 2/ +3^2. If P2 is atmospheric pressure, it may be neglected,
and
-sfcm
202. Sphere: Greater Pressure on Outside. — Here again t
changes to compression or reverses in sign, yielding
. 2j±fr r 2 * f'-p 2t 2 j'+p 2 r<?r-p*
3 ** 3 3 ^3
"-^mm ™
That fi shall be greater than zero requires that p 2 < i(2/+^i).
203. Diagrams of Stress. — Curves may be drawn to represent
the variation of p and / in the four preceding cases. They are
all hyperbolic, and, if r is laid off from the centre O on the hori-
zontal axis, each curve will have the vertical axis through O
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ENVELOPES.
213
for one asymptote, and for the* other a line parallel to the hori-
zontal axis, at a distance indicated by the first term in each value
of / or p. The four accompanying sketches show the various
curves. The values of / and f, the unit stresses in the material
at the interior and exterior, which correspond to the given values
of P\ and pz, are found at the extremities of the abscissas which
represent r\ and f2. The error which would arise from con-
sidering / as uniformly distributed is manifest. The dotted
— / V^l
\ Fig.8r
circles show the respective cylinders or spheres. Fig. 86 gives
the external and internal tensile stress for Pi in the interior of a
thick cylinder. Fig. 87 shows the distribution of compression
when the greater pressure is from without. Figs. 88 and 89
represent thick spheres under similar pressures.
204. Tank with Conical Bottom. — A water-tank of radius r
may be built with a conical bottom and be supported only at the
perimeter by a circular girder. The pressure of the water in
pounds per square inch at any point is p =0.434 X depth of point
below surface. In the cylinder the stress per unit of length on
a vertical joint is j^r; the stress on a horizontal joint is equal
to the weight of the sides lying above the joint and is generally
insignificant.
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214
STRUCTURAL MECHANICS.
Any horizontal joint in the cone such as A A of Fig. 90 must
carry a load, W, composed of the weight of water in the cylinder
whose base is the circle A A, the water in the cone ABA and
the weight of the metal shell of the same cone. The last item
is comparatively insignificant. As the cone, like the sides, is
built of thin plates, the stresses in the cone must always act
tangentially to the shell, so W +27&1 is the vertical component
of T\ 9 the stress per unit of length on the horizontal joint, and
_ W sec
27tT\
which varies from zero at B to a maximum at C.
To find the stress on a joint along an element of the cone
imagine a ring X>f slant height unity to be cut out by two hori-
zontal planes as shown in the figure. Substitute for the pressure,
p, which acts normally around the ring, the two components,
p tan and p sec 0. Of these the former acting along the ele-
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ENVELOPES. 215
ments causes no stress on an elemental joint while the latter
causes a tension in the ring of
^2= pf\ sec 0= pp.
It can be proved that n sec is the radius of curvature of the conic
section cut out by a plane normal to the element, B C, hence 72
is equal to the pressure into the radius of curvature as shown
in § 195. This tensile stress varies from zero at B to a maxi-
mum at C.
The total weight of the tank and contents is carried by
a circular girder, which in turn is supported by three or more
posts, consequently the girder is subjected to both bending
and torsional moments. At C the stress, T it is resolved into
vertical and horizontal components, the former of which is
W+27& and is the vertical load per unit of length of girder; the
. W tan , . L
latter is — — — which causes a compressive ring stress in the
„ W tan
girder of .
Example.— A circular tank, 40 ft. in diameter, has a conical bottom
for which 0= 45 . The depth of water above the apex is 60 ft. Weight
of cu. ft. of water, 62.5 lb. Sec 0= 1.414. Tension in lowest vertical
ring of sides is 40X0.434X20X12=4,170 lb. per lin. in. Tension in
radial joint of cone at A, half way up, is 50X0.434X120X1.414=3,680
lb. per lin. in. of joint. Same at C is 40X0.434X 240X 1.414= 5,000 lb.
per lin. m.
For tension on horizontal joint at A, half way up W=7rX
100(50+^X10)62.5. r 1 = iXioX53.33X62. 5 Xi.4i4=2 3 ,6oo lb. per
lin. ft.= i, 9 6olb. per lin. in. of joint. At C, T x = *X 20X 46.67 X 62. «?X
1.414=41,250 lb. per ft. = 3, 44 o lb. per lin. in. The vertical com-
ponent of T x at C is 41,250 + 1.414=29,200 lb. per ft. of girder. As
the horizontal and vertical components are equal, the compression in
the girder is 29,200X20=584,000 lb.
205. Tank with Spherical Bottom.— The stresses in the spheri-
cal bottom are found in the same way as in a conical bottom.
Any horizontal joint as A A carries the cylinder of water whose
base is the circle A A, the segment of water ABA, and that part
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2l6
STRUCTURAL MECHANICS.
of the shell below the joint. The volume of the segment is
na^r* — \a) if a=r'(i— cos0). If the weight on the whole joint
is W, the vertical component of the
stress per linear unit of joint is
W +270? sin and the stress is
W
Tl "2nr / sin 2 d'
The tension per unit of length on
any mer'dian joint is, by § 196,
The vertical load on the girder is
W ' +2TT per unit of length, and the
horizontal force applied to the girder
W
by the bottom of the tank is
J 2Tzr tan a
per unit of length, which causes a corn-
Fig. 91
pressive stress of
W
in the girder.
27: tan a
The stresses in the spherical bottom are smaller than those
in the conical bottom.
Example. — A circular tank, 40 ft. diameter and 40 ft. high, has a
spherical bottom for which a=45°- Then ^=28.3 ft. and the ex-
treme height=48.3 ft. Tension in radial joint at bottom=$X62.5X
28.3X48.3=42,645 lb. per linear ft.=3,554 lb. per in. of joint. Ten-
sion in radial joint at A, half way up, where 0=22$°, is 40,770 lb. per
ft. or 3,397 lb. per in. of joint. At C, tension is 35,350 lb. per ft., or
2,946 lb. per in. of radial joint. Tension in horizontal joint at A
is 41,760 lb. per ft., or 3,480 lb. per in., and at C is 39,220 lb. per ft,
or 3,268 lb. per in. of joint. Compression in circular girder= 554,700 lb.
206. Conical Piston. — The cone C B C of Fig. 92 represents
a conical piston of radius r, subtending an angle 28, with a normal
steam pressure, p f per unit of area applied over its exterior or
interior, the supporting force being supplied by the piston-rod
at B. The force in the direction of the rod on any section A A
of radius r\ is ^(r 2 — fi 2 ) which becomes at the vertex ^rr 2 , the
force on the piston-rod. This force will be compression on the
rod and tension in the cone, if p acts on the exterior of the cone,
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ENVELOPES.
217
and the reverse if p acts within the cone. The unit stress in the
metal of the cone at this section will be found by multiplying
this force by sec 0, and then dividing
by the cross-section, 2itr\t y in which / is
the thickness of the metal. The unit
stress on the circumferential section is
then
*-ip-
-r x 2 ) sec 0,
Fig. 92
which is a maximum at the piston-rod if / is constant.
The unit stress at A on a radial section is, by § 204,
/> 2 =y=— sec0.
When Pi is compressive, p 2 is tensile and vice versa.
Example. — Conical piston, Fig. 92. r=24 in., radius of rod=3 "*•»
0=69°. Thickness for fi=i7 in. is 1.5 in.; for 1^=8 in. is 1.9 in.
Steam pressure= 100 lb. per sq. in. Sec 0= 2.79. For r 1 = 1 7 in., pi =
ioo(24 2 ~i7 2 ) 2.79-^(2X17X1. 5) = i,57o lb. per sq. in.; p2=
100X17X2.79-^15 = 3,160 lb. per sq. in. For ^ = 8 in., £1 = 4,700 lb.
and £2=1,175 lb. For alternating stresses on steel castings these
stresses are satisfactory. See example § 175.
Pig. 93
207. Dome. — A dome subjected to vertical forces symmetri-
cally placed around its axis, such as its own weight, may be treated
as follows : C B C of Fig. 93 represents a meridian section of a
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2i8 STRCCTURAL MECHANICS.
dome, a hemisphere as shown, but the results to be deduced
are true for any surface of revolution about a vertical axis. If
a horizontal plane, A A, is passed through the dome to cut out a
circle of radius r x and all the weight from the crown to that sec-
tion is denoted by W, the stress in the shell per linear unit of
circumferential joint is
^ Wsecd
Ti .
21tTi
\ i is always compressive and is a maximum at the base.
To find the stress on a meridian section pass two horizontal
planes through the dome so as to cut out a thin ring of mean
radius r x . If the total load above the ring is W x and the total
vertical force supporting the ring is W2, the weight of the ring
is W2-W1. As the shell of the dome is thin the stresses in the
shell are tangential to the surface and the ring is acted upon
by a system of forces around its circumference as shown on the
right side of the figure. Resolve the forces into vertical and
horizontal components as shown on the left. Acting upon the
upper edge per unit of mean length of ring is the vertical corn-
el j t. ^ • , WitfiiOi „
ponent and the horizontal component . By sub-
stituting W2 and 02 for W\ and X the components acting on the
lower edge are found. The vertical components, together with
the weight of the ring balance among themselves, but there is
an unbalanced horizontal force of H = {W\ ctn X —W 2 ctn 2 )
which causes tension or compression in the ring depending on
whether it acts outward or inward. The stress in the ring is,
therefore, Hr x and its intensity per linear unit of joint, T2, is
found by dividing by the width of the ring measured along the
meridian. At the crown T2 is compressive and equal to T x ,
but it diminishes as A is taken lower and lower down and becomes
tensile in the lower part of the structure.
208. Resistance of a Ring to a Single Load. — When a ring is
acted upon by two equal and opporite forces as shown in Fig. 94,
the curve becomes flatter at A and sharper at B, showing that
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ENVELOPES.
219
bending moments of opposite sign are set up at these points.
From conditions of symmetry it is seen that each quadrant is
acted upon by vertical forces of \W together with an unknown
moment at each end. Imagine the quadrant to be removed from
1
hw
Fig. 04
the circle and horizontal levers to be attached at A and B so that
the forces can be moved horizontally such a distance as to cause
the actual moments existing at A and B. For equilibrium the
forces, when moved, must be applied in the same line, and the
moment at any section of the ring will be determined when the
line of application of the forces is fixed.
To fix that line the deformation of the arc must be considered.
1 M M
By § 88, ""=^7 whicfi becomes da^-^jds ft the angle between
the two radii of Fig. 44 is da. This equation gives the change
in the angle between two right sections ds apart, caused by the
bending moment, and is true for curved beams as well as straight.
To find the change between the two sections a distance s apart inte-
grate from zero to s. In the ring under consideration the tan-
gents at A and B remain horizontal and vertical respectively,
as seen from the condition of symmetry, hence the change in the
angle between right sections at A and B is zero and as E and /
Mds=o. M at any point C is JWXDC and
must be expressed in terms of r and 6 to be integrated. Then
iw 'f!
(a—rcor, 6)c!0=o;
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220 STRUCTURAL MECHANICS.
I (ad-r sin 0) =o = \aiz —r- 9
a=— =o.6366r.
The bending moments in the hoop can now be determined.
If a ring is acted upon by four normal forces 90 apart as is
the horizontal girder around the base of a water tank when the
posts are inclined, the bending moments at various points of the
girder may be found for each pair of forces independently and
the results added algebraically.
Examples. — 1. What is the net thickness required for a boiler shell
60 in. in diameter to carry 120 lb. steam pressure? What the gross
thickness allowing for riveting, and the size and pitch of rivets ?
J in.; -jV in.; J in.; 3 rows, 3 in. pitch.
2. What weight applied at top of circumference and resisted at
bottom ought a cast-iron pipe, 12 in. diam., J in. thick, and 6 ft. long,
to safely carry, if /= 12,000 lb.?
3. Determine the thickness of the cast iron cylinder of a 10-ton hy-
draulic jack to work under a pressure of 1,000 lb. per sq. in. if /= 4,000 lb.
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CHAPTER XIV.
PLATE GIRDERS.
209. I Beam. — A rolled beam of I section may be considered
composed approximately of three rectangles, — two flanges, each
of area 5i, and a web of area S w . The depth between centres
of stress of the flange sections may be denoted by h', which is
also very nearly the depth of the web. Then the resisting moment
of the two flanges will be fS f h', and that of the web, since M for
a rectangle is ifbh 2 , is j-\S w h f . The value for the entire section
will be
M=f(S f +lS w )h'.
Hence comes the rule that one-sixth of the web may be added
to one flange area in computing the resisting moment of an I
beam. The extreme depth of the beam ought not, however,
to be used for h\ The approximate distance between centres
of gravity of the flanges will answer, since it is a little short of
the true value for the flanges and a little longer than is correct for
the web.
2io. Plate Girder. — A portion of a plate girder and a section
of the same is shown in Fig. 95. Such a structure acts as a beam
and is designed to resist the maximum bending moments and
shears to which it may be liable. It may be loaded on top, or
through transverse beams connected to its web. It is used when
the ordinary sizes of I beams are not strong enough to resist
the maximum bending moment. As the flanges may be varied
in section by the use of plates where needed, as shown at the
right, there may be more economy of material in using a built
beam rather than a rolled one, if the required maximum section
is large.
221
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222
STRUCTURAL MECHANICS.
4 +
t 4
+ 4
1 4-
4-4 1
^ ^
r* + 1
<■ i
f +.
H* 4
+ t
f* ■ +
f +
h 4
4- +
.+ • -t
4 i
K 4
> ♦
*. t
+ +
I+" +
t t
f J t
+ t
V *
t 4
t .-H
4 t
4-' *
+' t
f 1
-T 4 4
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PLATE GIRDERS. 223
The web A is made of sufficient section to resist the maxi-
mum shear, and the rest of the material is thrown into the flanges,
B, where it will be farthest removed from the neutral axis and
hence most efficient in resisting bepding moment As the thick-
ness of the web plate is usuallywS^^to one-fourth inch, and
in girders of any magnitude to three-eighths inch, as a minimum,
it appears that the material in the web practically increases with
the depth of the girder. As the stress in either flange multiplied
by the distance between the centres of gravity of flanges resists
the bending moment, the material in the flanges decreases as the
depth increases; hence the most economical depth is that which
makes the total material in the web, including the splice plates
and stiffeners, as near as may be equal to that in the two flanges.
Depth of beam contributes greatly to stiffness, when a small
deflection is particularly desirable, and the depth may in such
a case be so great as to make the web the heavier.
211. Web. — Some engineers apply the rule of § 209 to a plate
girder and consider the flange section to be increased by one-
eighth of the web section, the fraction one-eighth being used
instead of one-sixth because of the weakening of the web by rivet-
holes; but the more commonly received practice is to consider the
flanges alone as resisting the bending moment at any section
and the web as carrying all the shear uniformly distributed over
its cross-section as shown in § 73. If the web is required to
resist its share of the bending moment, web splices must be
designed to transmit the stresses due to bending as well as those
due to shear. Although web plates are generally made of the
thinnest metal allowable, the designer is prevented from using
too thin a web plate by the fact that thin plates offer so little
bearing resistance to the rivets. If the plate is too thin it will
generally be impossible to put in a sufficient number of rivets
to connect the web to the flange.
212. Flanges. — If the maximum bending moments are com-
puted for a number of points in the span of the girder, they can
be divided by the allowable unit tensile stress times the effective
depth (that is, the distance between centres of gravity of flanges)
to give the required net sections of the tension flange at those
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224 STRUCTURAL MECHANICS.
points. The compression flange is usually made of the same
gross section as the tension flange. The deduction for rivet-holes
in the latter, which is not necessary in the former, compensates
for the slightly lower unit stress allowed for compression. The
compression flange must be wide enough not to bend sideways
like a strut between the points at which it is stayed laterally. As
the web checks such lateral flexure in some degree, and as the
flange stress varies from point to point, it is impracticable to
apply a column formula. In practice the unsupported length of
plate is not allowed to exceed a certain number of times the width
of the flange: for railroad bridges different specifications give
from 12 to 20; for girders carrying steady load only, as high a
number as 30 may be used.
The selection of the plates and angles to give the required
flange section is largely a matter of judgment. The angles must
be large enough to support well the compression flange-plates
and to be able to transmit the increments of stress from the web
to such flange-plates. Hence the area of the flange-angles should
be a considerable portion of the total flange section. For railway
girders it is often specified that the section of the angles shall
be at least one-half that of the whole flange, or that the largest
angles procurable shall be used.
213. Length of Flange-plates. — As the bending moment varies
from point to point in a girder, the required flange section will
vary in the same way, being, in general, greatest at the half span.
If the section at that point is made up of two angles with flange-
plates, the girder may be made to approximate to a beam of
uniform strength by running the flange-plates only such distance
from the centre as they are needed.
Inspection of the necessary sections will show how far from
the two ends of the girder, as IK, the flange-angles, with rivet-
holes deducted, will suffice for the required flange section. From
K to the corresponding distance from the other abutment the first
plate must extend. A reasonable thickness being used for that
plate, with a deduction for rivet-holes in the tension flange, it
can again be seen where a second plate will be needed, if at all.
This determination can be neatly made on a diagram of maxi-
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PLATE GIRDERS.
225
mum moments, or on the similar diagram showing the necessary
sections. Extend the plate either way a small additional dis-
tance, to relieve the angles and assure the distribution of stress
to the plate. The thicker plate, if there is any difference in thick-
ness, should be placed next to the angles.
When the girder carries a uniformly distributed load the
^*—>
r"-^
-o-
I
£
o-
to X
i i
|'i!
-iS —
Y
'' \
A
/ s \
/ 1 \
, r
Fisr
re
Fig. 97
required flange areas vary as the o dinates to a parabola (Fig. 96)
and the length of any flange-plate is given by
x =Ns'
in which / is the length of the girder from centre to centre of
base-plates, S the net flange area required at the centre, and s
the net area of the plate whose length is desired plus the area
of such plates as may lie outside it.
If the girder is so long that the plates or angles must be
spliced, additional cross-section must be supplied by covers at
the splices, with lengths permitting sufficient rivets to transmit
the force. Even compression joints, though milled and butted
together, are spliced in good practice. The net area of the cover-
plate and splice angles should be equal to that of the largest
piece spliced. Only one piece should be cut at any one section,
and enough lap should be given for the use of sufficient rivets
to carry the stress the piece would have carried if uncut.
214. Rivet Pitch. — If a strip of flange (Fig. 97) of a width
equal to the pitch of the rivets connecting the flanges to the web
is cut out by two imaginary planes, shearing forces, F, act on the
two sides forming a couple with an arm equal to the pitch. Under
the usual assumption that the web carries shear only, there are
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226 STRUCTURAL MECHANICS.
no other forces acting on the planes of section, and the only forces
which can keep the strip from rotating are those supplied by the
two flange rivets. Hence by equating the two couples the pitch
can be found.
_. . rivet value X depth
Pitch = r *— .
shear
As the flange-angles are supposed to be fully stressed at the
point where the first flange-plate begins, the increments of flange
stress coming out of the web must pass through the flange-angles
and into the flange-plate. The rivets connecting the flange-angles
and the flange-plates must therefore resist the same stress as do
the rivets connecting the web plate and the flange-angles, and the
above formula applies to both, although the rivet values in the
two cases will be different. The rivets through the web are in
bearing (or double shear if the web is thick), while those in the
flange-plate are in single shear and occur in pairs. But practi-
cally the pitch in one leg of an angle must be the same or an even
multiple of the pitch in the other, so that the rivets may be stag-
gered.
Make the pitch of rivets in inches and eighths, not decimals;
do not vary the pitch frequently, and do not exceed a six-inch
pitch, so that the parts may be kept in contact. If flange-plates
are wide, and two or more are superimposed, another row of
rivets on each side, with long pitch, may be required to insure
contact at edges. Care must be taken that a local heavy load
at any point on the flange does not bring more shear or bearing
stress on rivets in the vertical legs of the flange-angles than allowed
in combination with the existing stress from the web at that place
and time.
Webs are occasionally doubled, making box girders, suitable
for extremely heavy loads. The interior, if not then accessible
for painting, should be thoroughly coated before assembling.
If the web must be spliced, use a splice plate on each side
for that purpose, having the proper thickness for rivet bearing
and enough rivets to carry all the shear at that section; there
should be two rows of rivets on each side of the joint.
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PLATE GIRDERS. 227
215. Stiffeners. — At points where a heavy load is concentrated
on the girder, stiffeners, C, consisting of an angle on each side,
should be riveted to throw the load into the web and to prevent
the crushing of the girder. They should, for a similar reason,
be used at both points of support, D. Such stiffeners act as
columns and may be so figured, but as the stress in them varies
from a maximum at one end to zero at the other, a good rule
is to consider the length of the column equal to half the depth
of the girder.
Since the thrust at 45 to the horizontal tends to buckle the
web, and the equal tension at right angles to the thrust opposes
the buckling, it is conceivable that a deep, thin web, while it
has more ability to carry such thrust as a column or strut than it
would have if the tension were not restraining it, may still buckle
under the compressive stress ; and it is a question whether stiffeners
may not be needed to counteract such tendency. They might
be placed in the line of thrust, sloping up at 45 from either
abutment, but such an arrangement is never used. They are
placed vertically, as at C, and spaced by a more or less arbitrary
rule.
A common formula is : The web of the girder must be stiffened
if the shear per square inch exceeds
d
10,000 -75 j,
where d= clear distance between flange-angles, or between stiffeners
if needed, and / = thickness of web. Another rule calls for
stiffeners at distances apart not greater than the depth of the
girder, when the thickness of the web is less than one-sixtieth
of the unsupported distance between flange-angles. There is
no rational method of determining the size of stiffeners used
only to keep the web from buckling. The usual practice is about
this : make the outstanding leg of the angle 3 \ inches for girders
less than 4 feet deep, 4 inches for girders 5 feet deep, and 5 inches
for girders over 7 feet deep.
Experience appears to show that stiffeners are not needed at
such frequent intervals as the formula would demand. An
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228 STRUCTURAL MECHANICS.
insufficient allowance for the action of tension in the web in
keeping the compression from buckling it, is probably the cause
of the disagreement.
Interior stiffeners may be crimped at the ends, or fillers may
be used under them to avoid the offset. End stiffeners and
stiffeners under concentrated loads should not be crimped; they
should fit tightly under the flange that the load may pass in at
the ends.
Example. — A plate girder of 30 ft. span, load 3,000 lb. per ft.,
/= 15,000 lb. per sq. in., ^=90,000 lb., and Af m a*. == iW^
= 4,050,000 in.-lb. Assume extreme depth as 42 in., effective depth,
39 in. Net flange section at middle = 4,050,000 -*- (39 • 15,000) = 7 sq. in.
A f-in. web 42 in. deep will have 15! sq. in. area. Two flanges,
each 7 sq. in. net + allowance for rivet-holes, will fairly equal the web.
Use f-in. rivets.
Let the flange-angles be2— 4X3X1 in. = 4.96 sq. in. Deduct 2 holes,
fX}=o.66. Net plate= 7 — 4.3 = 2.7 sq. in. A plate 9Xf=3f sq. in.;
deduct two holes=o.66, leaving 2.71 sq. in. Two angles and plate,
gross sec ^=4.96+3.37=8.33 sq. in. Resisting moment of net
section of angles=4-3X 15,000X39= 2,515,500 in.-lb. Such a bending
moment will be found at a distance x from either end, given by Pix—
i'3,ooo» 2 = 2,515,500. #=5.9 ft. .". Cut off the plate 5 ft. from
each end.
Shearing value of one f-in. rivet at 10,000 lb. per sq. in. =4,400 lb.
Bearing value in f-in. plate at 20,000 lb. = 5,600 lb. Max. shear in
web =45,000 lb. Pitch for flange-angles, since bearing resistance is
less than double shear, =5,600 -39 -*- 45,000= 4.85 =4$ in. Make 3-in.
pitch for 2 ft., then 4|-in. for 6 ft., then 6-in. pitch to middle. Rivets
in end web stiffeners, 45,000-7-5,600=9. Max. shear in web= 45,000
d
-7- (42 •§) = 2,780; 10,000— 75-- =2,800, since (£=42— 6. No other
1
stiffeners needed. By the other rule 36-5-1=96, and stiffeners are
needed.
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CHAPTER XV.
SPRINGS AND PLATES.
216. Elliptic Spring. — The elliptic spring is treated in § 97.
If the load on the spring is W and the span is /, the deflection is
The fibre stress is
1 WP
V ~i6EIo
6Mp 3 Wl
'~b Q h 2 2 b Q h 2 '
and the work done upon the spring is \Wv, which is equal to the
resilience of the spring. Hence
1 W 2 P 1 p
Resilience = — =rr- = — -= • volume,
32 EIq 6 E
217. Straight Spring. — If a beam of uniform section, fixed at
one end, has a couple or moment applied
to it (Fig. 98) in place of a single transverse
force, it will, as shown in § 89, bend to the
arc of a circle. The deflection will be, if
/ is the length of the beam,
Kg.tt
v=
MP }P
2EI 2Ey{
The work done by the rotation of a couple is the product
of its moment by the angle through which it turns. For a deflec-
229
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230 STRUCTURAL MECHANICS.
tion, dv, the free end of the spring turns through an angle 2dv -*-/;
hence
Resihence=2y ydv-.-p-y ,^_.^_-_._
For a rectangular section iA, these quantities become
6A// 2 /Z 2
V ~Ebh?~~Eh'
if 2 1 P
Resilience — t- • -f • ft W = t- • -^ • volume,
O jCr O JCr
For a circular section the number 6 in the last expression
will be replaced by 8.
218. Coiled Spring. — In practice the rectangular or cylin-
drical bar is bent into a spiral and subjected to a couple which,
as a couple can be rotated in its plane without change, acts equally
at all sections of the spring. The developed length of the spiral
is/.
219. Helical Spring. — A cylindrical bar whose length is / and
diameter d, when fixed at one end and subjected to a twisting
moment T— Pa at the other, if the elastic limit is not exceeded,
32 77
by § 84, is twisted through an angle 0= f^i- The work ex-
pended in the torsion is
From § 84, = 7v> and therefore
Resilience = --^ /= — 7; -volume.
4 C 4 4 C
2 4 2 P
HC=-£ and ?i=-/, work =—•-»• volume, while for flexure,
1 /2
as shown in the preceding section, work = T—p' volume, a smaller
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SPRINGS AND PLATES. 23 I
quantity; hence a spring of given weight can store more energy
if the stresses are torsional than if they are bending.
If this bar is bent into a helix of radius, a, and the force, P,
is applied at the centre in the direction of the axis of the cylinder,
the moment, Pa, will twist the bar throughout its length. Then
i6Pa n n(P
The deflection of the spring is v=aO, since, as the force P de-
scends, the spring decends, and the action is the same as if the
spring remained in place and the arm revolved through an angle
0. The force P is too small to cause any appreciable compression
(or extension) of the material in the direction of its length.
32 Pa 2 l_2al q x 47tna 2 q x
if n— number of turns of the helix and l = 27tan.
If the section of the spring is not circular, substitute the
proper value of q\ or the resisting moment from § 85. If the
/ d'*\
rod is hollow, multiply the exterior volume by ( 1— -»). For a
square section and a given deflection, P will be about 65 per cent,
of the load for an equal circular section. C for steel is from
10,500,000 to 12,000,000.
Example. — A helical spring, of round steel rod, 1 in. diameter,
making 8 turns of 3-in. radius, carries 1,000 lb.
16-1,000-3-7 4-22-8-qi5,273
?i = — ' = 15,273. v=- J =1.15 in.
** 22 D lo 7-1-12,000,000 D
220. Circular Plates. — The analysis of plates supported or
built in and restrained at their edges, and loaded centrally or
over the entire surface, is extremely difficult. The following
formulas from Grashofs "Theorie der Elasticitat und Festig-
keit " may be used. The coefficient of lateral contraction is
taken as J, or m=*4.
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23 2 STRUCTURAL MECHANICS.
I. Circular plate of radius r and thickness /, supported around
its perimeter and loaded with w per square inch.
/ x =unit stress on extreme fibre in the direction of the radius,
at a distance x from the centre;
/„ = unit stress perpendicular to the radius, in the plane of the
plate, at the same distance x from the centre.
/*=f8?(V' 2 -3*°); /.-fjc*" 2 -* 2 )-
t , t( N 117 wr 2 i&our*
h = Ivma*.(t0rX = 0)=— sir . Vq = — —
For the same value of /, the maximum stress is independent
of r, provided the total load vmr 2 is constant.
II Same plate, built in or fixed at the perimeter.
t 45 " W , o o\ t 45 w o ox
At the centre, /*=/ y . At the circumference /„ is zero, and /*
is maximum.
1 _45 ^f „, _ 45_^*
III. Circular plate supported at the perimeter and carrying
a single weight W at the centre. Loaded portion has a radius r .
/i= ^i|( Iog i_i). /r iijL: + i),
' 32^ t 2 \ b X 5/' /W 327T/ 2 \ & # 5/
These expressions become maxima for #=ro, and the second is
the greater.
117 Wr 2
For values of r+r = 10, 20, 30, 40, 50, 60,
/max. =i-4 i«7 i-9 2.0 2.1 2.2 W-f/ 2 .
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SPRINGS AND PLATES. 233
If ro=o, the stress becomes infinite, as is to be expected,
since W will then be concentrated at a point, and the unit load
becomes infinitely great. It is not well to make tq very small.
IV. Same plate, built in or fixed at the perimeter.
, 45 W/ r \ 45 W r
The maximum value of / is /„, for x=ro.
45 Wr 2
For values of r+r = 10, 20, 30, 40, 50, 60,
/max. =1.0 1.3 1.5 1.6 1.7 1.8 W+P.
221. Rectangular Plates. — The problem of the resistance of
rectangular plates is more complex than that of circular plates.
Grashof gives the following results :
V. Rectangular plate of length a, breadth 6, and thickness /,
a>6, built in or fixed at edges and carrying a uniform load of
w per square inch.
b 4 -wa 2 a 4 -tab 2
' 2 (a 4 + 6 4 )/ 2 ' /0 2(a 4 +fr*)/2'
The most severe stress occurs at the centre in the direction fc,
that is, on a section parallel to a. If
wa 2
a 4 fr* a/
The deflection at the centre is v = ttta — f^» and for a
a 4 + 4 32Et 6
. 7fa 4
square plate, ^.
VI. Plate carrying a imiform load of w per square inch and
supported at rows of points making squares of side a. Fire-
box sheet with staybolts.
i$wa 2 ' iq wa 4
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234 STRUCTURAL MECHANICS.
Navier gives formulas for rectangular plates which are sup-
posed to be very thin. Approximate values from those formulas
are as follows :
VII. Rectangular plate, as in V, but supported around the
edges.
a 4 b 2 w a 4 b 4 w
/=°-92 (a2 + 62)2 J 2 J oo =0-19^2 + ^2 Et>
VIII. Rectangular plate, supported at edges and carrying a
single weight W at centre.
I O a3l) W A aW W
< = 2 - 28 { aZ + b*)*T* ; " 0= ° 4 VW1/ 3,
. For the same total load, / is independent of the size of the
plate, provided the ratio a to b and the thickness are unchanged.
Example. — A steel plate 36 in. square and i in. thick, supported at
edges, carries 430 lb. per sq.ft., or 3 lb. per sq. in. /=o.92-£-36-36-3*i6
= 14,300 lb.
^019 3 6*- 3 -4 3
4 30,000,000
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CHAPTER XVI.
REINFORCED CONCRETE.
222. Reinforced Concrete. — As concrete has small tensile
strength and is likely to crack when built in large masses, it can
be reinforced to advantage by steel bars or wire netting imbedded
within it. This form of construction is much used and is espe-
cially applicable to beams and slabs, in which the steel is placed
near the tension edge. By its use a great saving of material
in masonry structures can often be effected, since the strength
of the structure can be depended upon, rather than its weight.
Among its advantages for buildings may be mentioned the fact
that it is fireproof and that the metal is protected from rust.
The expansion and contraction of the two materials from changes
of temperature are so nearly alike that heat and cold produce
no ill effects. When used in beams of any considerable span,,
there is the disadvantage that the dead load is a large proportion
of the total load.
To compute the strength of a structure composed of two
materials which act together, the modulus of elasticity of each
material must be known, and it is here that the chief difficulty
in computing the strength of reinforced concrete members lies.
The modulus of elasticity of concrete is uncertain; it not only
depends upon the composition of the concrete, but, for a given
concrete, varies with the age, while for a particular specimen the
stress deformation diagram is not a straight line as for steel, but
the ratio of the stress to the deformation decreases with the load.
The concrete takes a permanent set, even for small loads, in
consequence of which a reinforced-concrete beam, after being
released from a load, is in a condition of internal stress.
235
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236
STRUCTURAL MECHANICS.
Considering, then, our lack of knowledge of the exact stresses
to be expected in the concrete, it is best to adopt as simple a
method of computation as is reasonable.
223. Beams. — Beams may be figured according to the common
theory of flexure, and the following assumptions will be made:
1. The steel and the surrounding concrete stretch equally.
2. Cross-sections, plane before bending, are plane after bending,
3. The modulus of elasticity is constant within the working
stress.
4. The tension is borne entirely by the steel. As the tensile
strength of concrete is low and as a crack in the beam would
entirely prevent the concrete from resisting tension, this is a
proper assumption to make and is on the side of safety.
5. There is no initial stress on the section.
Neu tral
Axis
< — b — >
— • — •—
r
FJfiT-90
From these assumptions it follows that the stresses acting
on the cross-section are as shown in Fig. 99.
Let E 8 = modulus of elasticity of steel;
E c = " lt " " concrete;
6 = hi 8 — hi C ;
/ a = unit stress in steel;
} c = " " " concrete at extreme compression fibre;
X = deformation of fibre at unit distance from the neutral
axis, between two sections originally 'unit distance
apart;
kbh = sectional area of steel;
y= distance from neutral axis to extreme compression
fibre;
« = a numerical coefficient.
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REINFORCED CONCRETE. 237
By § 10,
f.-XE.(h-y), f e =*E c y,
or the ratio between the unit stresses is
h~E e y ~ e y (i;
As the total tension on the cross-section must equal the total
compression, for a rectangular cross-section
\byU = kbh} a (2)
Combining (i) and (2) gives
u _ y _ c h -y %
} c 2kh y
Solving this equation for y gives the location of the neutral axis,
y = (-ek+Vc 2 k' : + 2ek)h=ch (3)
This equation shows that for beams made of a given quality
of concrete the location of the neutral axis depends only upon
the percentage of reinforcement.
The resultant of the compressive stresses on the cross-section
is applied at a point Jy from the top of the beam and together
with the tensile stress in the steel forms a couple whose arm is
h—\y. The moment of this couple, which is the resisting moment
of the beam, is
M = \byUh~\y) (4)
= kbhtth-ly),
the first expression giving the moment in terms of the stress in
the concrete and the second in terms of the stress in the steel.
If the value of y from (3) is substituted in (4) there results
M = \di-lc)j J>h 2 =n e }M 2 (5)
= k(i-ic)l a bh 2 =n a } a bh 2 .
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338
STRUCTURAL MECHANICS.
As long as } c and /, are the unit stresses in the concrete and in
the steel, the two forms of (5) must be equal, but if j c and / a are
arbitrarily chosen working stresses, that form of (5) must be used
in designing which will give the smaller value of M. Therefore
/. c
use n e when wj c <w,/ a or when t> ~r- The actual fibre stresses,
however, can be made to assume any given ratio by changing the
proportion of steel reinforcement, and that proportion can be
found by eliminating y from (1) and (2) with the result
*-i
eft
2l. 2 + e}cf. J.l).
f.(H
(6)
Equation (5) can be readily applied to the design of beams
by tabulating the values of n c and n 8 for different values of k.
k
*-8.
e— 10.
e— 12.
c
««
n%
c
1U
n,
c
tie
n$
.003
.004
.005
.006
.007
.008
.010
.012
.014
.016
.018
.020
196
223
246
266
284
300
328
353
375
394
4ii
428
092
103
"3
121
128
135
146
155
164
171
177
183
.00280
.00370
.00459
.00546
.00633
.00720
.00891
.217
.246
.270
.292
•3"
.328
.358
.384
.407
.428
• 447
.463
.101
•113
.123
•132
.140
.146
•157
.167
.176
.183
.190
.196
.00278
.00367
.00455
.00542
.00628
.00713
.00880
235
265
291
314
334
353
384
411
435
457
476
493
I08
121
131
I4O
I48
155
167
177
186
194
200
206
00277
00368
00452
00537
00622
00706
00872
When reinforced concrete beams are tested to destruction, they
sometimes fail by the opening of diagonal cracks which seem to
follow in a general way the lines of principal stress. To prevent
that mode of failure most designers provide some form of rein-
forcement in a vertical plane. Stirrups or loops of wire lying
in planes of section and spaced at intervals less than the depth
of the beam are often employed, but no satisfactory method of
determining their size and spacing has been proposed.
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REINFORCED CONCRETE. 239
Another mode of failure to be guarded against is the slipping
of the steel in the surrounding concrete. That the beam may not
be weak in this respect, the change of stress in the bars, between
two sections a short distance apart (say one inch), must not
exceed the area of the surface of the bars between the two sections
multiplied by the safe unit adhesive stress of concrete to steel.
This change of stress between two sections one inch apart is
F+h(i— $c), since the shear in the beam measures the change
of bending moment at any section, as is shown in § 56. Rods
which have been roughened or corrugated may be used to diminish
the likelihood of slipping.
Example. — Design a beam of 16 ft. span to carry an external load
of 500 lb. per ft. if Jc=S°° lb., / 5 = 16,000 lb., and e=io, assuming
the beam to weigh 300 lb. per ft. 3f=£-8oo- 16-16-12=307,200
in.-lb. If each material is to be stressed to its limit, the proportion of
reinforcement needed is 10-7-2-32-42=0.00372, or say 0.4%. Then
n c =o.ii3 and M 2 = 307,200 -7- (0.113X500) = 5,430. 6=i2j in., /t=2i
in., and the area of steel is 0.004X12^X21 = 1.03 sq. in., or say 4—^
in. rounds. If there are 2 in. of concrete below the rods, the beam
weighs 150X121x23-7-144=284 lb. per ft.
If 0.8% of steel is used, instead of 0.4%, bh 2 = 4210. b= 10J in., h= 20
in., and 1.68 sq. in. of steel is needed; use 1 — $ in. and 2— J in. rounds.
Adhesion between concrete and steel. F max .= 800X8 =6,400 lb.
For first beam 6,400-7-21(1 — $X 0.246) =332 lb. Superficial area of
4- 1 V rounds=4-A , V = 7-°7 sq. in. 332-f- 7.07=47 lb. per sq. in.
For second beam 359-7-7.86=46 lb. per sq. in.
224. Columns. — Reinforced concrete columns are built with
steel rods embedded parallel to and spaced symmetrically about
the axis. The rods should be tied together by wire or bands at
intervals not greater than the diameter of the column. A common
design is a square section with the rods near the corners. As
the ratio of length to breadth is generally small, it is usual to
design such columns as short blocks.
The ratio between the intensity of the stress in the steel and
in the concrete is
U~E C - '
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240 STRUCTURAL MECHANICS.
and the total load the column may carry is
P = /c5(i -k) +j,kS = j c S[i +*(e-i)],
in which S is the total area of cross-section of the column.
With the usual working stresses and ratio of E 9 to E c employed
in designing, the greatest allowed unit stress in the reinforcement
is so small that economy requires the amount of steel to be
reduced as much as reasonable, hence the load P will be but
slightly greater than jcS.
225. Safe Working Stresses. — The following safe unit stresses
may be used in buildings:
Concrete, bending 500 lb. per sq. in.
* ' direct compression 350 ' l lt il "
Steel, tension 16,000 " " " V
Adhesion of concrete to steel 50 " " " '
E a +E c = 10 is a fair average value; the ratio may vary con-
siderably without materially affecting the proportions of the
beam.
The weight of reinforced concrete is about 150 lb. per cu. ft.
The concrete should be rich; one part cement, two, sand;
four, broken stone is a good mixture, the stone being broken to
pass through a f-inch ring. It should be mixed wet and placed
with great care to insure the proper bedding of the steel.
The proportion of steel reinforcement to use in beams is a
question of economy, which is most easily solved by designing a
number of sections of the same ratio of b to h, but with different
values of k y and figuring the cost per foot of each. A considerable
variation in k affects the cost but slightly, for ordinary prices of
concrete and of steel. The percentage of reinforcement is usually
between \ and it; £=0.007 is an average value.
The concrete lying below the reinforcing rods serves merely
to protect the steel. For fireproofing two inches is sufficient.
Steel thoroughly covered with concrete does not rust, and bars,
which were covered with rust when placed in concrete, have been
found to be bright when removed after some time.
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INDEX.
Angle of repose, 181
Annealing, 29
Ashlar, 30
Axis, neutral, 56, 61, 79
Bauschinger's experiments, 154
Beams, 2, 51, 87, 107
bending moments, 41, 44
cantilever, 45
Clapeyron's formula, 1 18
column and beam, 149
continuous, 116
curved, 60
deflection, 87
elastic curve, 87
fixed, 107
flexure of, 87
flitched, 102
I beams, 69, 173, 221
impact, 103
inclined, 60
modulus of rupture, 59
moving loads on, 48
neutral axis, 56
oblique loading, 78
reactions, 38
reinforced concrete, 236
resilience of, 103
restrained, 107
sandwich, 102
shaft and beam, 84
shear, external, 43
shear, internal, 65
slope, 88
stiffness, 89
stresses in, 54, 65, 179
three-moment theorem, 1 16, 122
tie and beam, 130
timber, 68
torsion on, 84
uniform strength, 63, 97
work, internal, 104
Bending and compression, 149
and tension, 130
and torsion, 84
Bending moment, see Moment, 41
Bessemer process, 25
Blocks in compression, 15, 137
Boilers, 189, 203
rivets, 198
working stresses, 166
Bricks, 31
Bridges, shear in panel, 52
working stresses, 162, 164
Buildings, working stresses, 165
Burnettizing, 21
Cantilevers, 45, 47
Carbon and iron, 22
Cast iron, 22
properties of, 23
working stresses, 166
Cement, 33
Centrifugal force, 133
Clapeyron's formula, 118
Clay, 31
Coefficient, see Modulus.
Columns, 2, 137
beam and column, 149
deflection, 142
designing, 146
eccentric load, 148
ends, fixed or hinged, 146
Euler's formula, 142
flexure, direction of, 141, 146
Gordon's formula, 144
ideal column, 142
Jacing-bars, 151
pin ends, 147
radius of gyration, 146
Rankine's formula, 144
reinforced concrete, 239
short. 137, 145
straight-line formula, 147
241
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242
INDEX.
Columns, swelled, 148
timber, 162
transverse force on, 149
working stresses, 162
yield-point, 141
Combined stresses, 3
bending and compression, 149
41 ** tension, 130
" " torsion, 84
tension and torsion 133
Compression, 3, 15, 137
bending and, 149
eccentric load, 137
granular materials under, 15
Concrete, 35, 235
Concrete-steel, 235
Cone, stresses in, 213, 216
Connecting-rod, 132
Continuous beams, 116
Cooper's lines, 190
Crank, 84
Curvature of beams, 88
Curve, elastic, 87
stress-deformation, 9
Cylinders, thick, 206
thin, 203
Deflection of beams, 87
simple, 89
restrained, 107
uniform strength, 97
Deflection of columns, 142
Deformation, 6, 184
Distortion, 7, 186
Dome, 217
Ductility, 6, 17, 26, 29
Earth pressure, 181
Eccentric load, 127, 137, 148
Elastic curve, 87
Elastic limit, 9, 154
Elasticity, modulus of, £, 6
cast iron, 23
concrete, 235, 240
steel, 26
stone, 30
timber, 22
wrought iron, 24
Elasticity, shearing modulus of, C, 7,
186
Ellipse of stress, 174
Elongation, work of, 8, io, 13, 27
Envelopes, 203
Equilibrium, conditions of, I
Euler's formula, 142
Eyebars, 131, 134
Fatigue of metals, 154
Flexure, common theory of, 54
Girders, see Beams.
Girder, plate, see Plate girder, 221
Gyration, radius of, 71
Hooks, 129
Hoops, 210
I beam, 69, 173, 221
Impact, 159
Inertia, moment of, 57, 71
product of, 76
Iron, cast, 22
malleable, 28
wrought, 24
Joints, masonry, 138
riveted, 195
Lattice bars, 151
Launhardt-Weyrauch formula, 156
Lime, 32
Lines of principal stress, 180
Linseed oil, 36
Loads, dead and live, 159
eccentric, 127, 137, 148
sudden application, 14
wheel loads on beam, 49
Machinery, working stresses, 166
Manganese in steel, 26
Masonry, 30
working stresses, 166
Materials, 19
Middle third, 138
Modulus of elasticity, see Elasticity, 6
of resilience, 13
of rigidity, 186
of rupture, 59
section, 57
Moment, bending, 41
maximum, 44
on p'ns, 200
position of load for maximum, 49
sign of, 41
Moment of inertia, 7 1
Moment of resistance, 57
oblique loading, 78
Moment, torsional, 81
Mortar, 32
Neutral axis, 56, 61, 79
Nuts, 134
Oblique load on beam, 78
Open -hearth process, 25
Paint, 36
Parallel rod, 133
Pedestals, 166
Permanent set, 8
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INDEX.
^43
Phosphorus in steel, 26
Pier-moment coefficients, 121
Pig iron, 23
Pins, 109, 200
distribution of shear in, 67
friction, 147
working stresses, 163
Pipes, 203
Piston, conical, 189, 216
Plaster, 33
Plate girder, 221
stresses in web, 69, 173
Plates, resistance of, 231
Polar moment of inertia, 71
Portland cement, 34
Posts, see Columns, 137
Power, shaft to transmit, 83
Principal stresses, 170
lines of. 180
Product of inertia, 76
Puddling furnace, 24
Pull and thrust at right angles, 172
Punching steel, effect of, 28
Radius of gyration, 71
Rafter, 60
Rankine's column formula, 144
Rankine's theory of earth pressure, 181
Reactions of beams, 38
Rectangular beams, 58
Repose, angle of, 181
Resilience, definition, 13
modulus of, 13, 103
of bar, 13
of beam, 103
of springs, 229
Resisting moment, see Moment, 57
Restrained beams, 107
Retaining wall, 182
Rigidity, modulus of, 187
Ring under normal pressure, 205
under single load, 218
Rivets, 192
plate girder, 225
steel for, 26
working stresses, 163
Rollers, 163
Rubble. 31
Rupture, modulus of, 59
Safe working stresses, 153
Sandwich beams, 102
Screw threads, 134
Secondary stresses, 158, 197
Section modulus, 57
Set. permanent, 8
Setting of cement, 34
Shafts, 81
working stresses, 166
Shear, 3, 43
Shear, deflection due to, 105
derivative of bending moment, 45
distribution on section of beam, 65
modulus of elasticity. C, 186
position of load for maximum, 48, 50
sign of, 43
timber beams, 68
two shears at right angles, 169, 172,
186
work of, 105
Shearing planes, 176
Sign of bending moment, 41
compression and tension, 4
shear, 43
Silicon in iron, 23
Slope of beam, 88
Spangenberg's experiments, 153
Sphere, stresses in, 205, 211
Splices, 195
in plate girder, 223, 225, 226
Springs, 98, 229
Steel, 25
shearing and punching, 28
structural, 26
tool, 28
working stresses, 162
Steel concrete, 235
Stiffness of beams, 89
Stiffners, 227
Stirrups, 238
Stone, 29
Straight line formula, 147
Strain, see Deformation.
Strength, beams of uniform, 6$, 97
cross-sections of equal, 62
ultimate, 1 1
Strength of cast iron, 23
steel, 26
timber, 22, 160
wrought iron, 24
Stresses, 2, 167
alternating, 154
conjugate, 170
distribution on section of beam, 54.
65. *79
ellipse of, 174
internal, 2, 167
lines of principal, 180
principal, 170
reversal of, 155, 162
secondary, 158, 197
sign of, 3
unit, 4. 167
working. 160. 240
Stress deformation diagram, 9
Struts, see Columns, 2, 137
Sudden loading, effect of, 14
Sulphur in steel, 26
Tanks, 213
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244
INDEX.
Tempering, 26, 28
Tension, 3, 127
bending and, 130
connections, 133, 196
eccentric load, 127
torsion and, 128, 133
Three-moment theorem, 116, 122
Tie, 2, 127
and beam, 130
Timber, 19
column formulas, 161
modulus of elasticity, 22
shear in beams, 68
strength of, 22
working stresses, 160
Torsion, 81
bending and, 84
resilience of, 230
tension and, 133
twist of shaft, 83
Trees, growth of, 19
Twist of shaft, 83
Ultimate strength, 11
Uniform strength, beams of, 63, 97
Unit stresses, 4
Varnish, 36
Varying cross-section, 12
Volume, change of, 183
Wall, retaining, 182
middle third, 138
Web of plate girder, 223
stresses in, 69, 173
Welding, 26
Wheel loads. 49
Wohler's experiments, 153
Wood, 19
Work of elongation, 8, 10, 13, 27
flexure, 104
shear, 105
springs, 229
Working stresses, 153
reinforced concrete, 240
Wrought iron, 24
Yield point, 10
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Furman's Manual of Practical Assaying 8vo, 3 00
* Getman's Exercises in Physical Chemistry nmo, 2 00
GilTs Gas and Fuel Analysis for Engineers 12 mo, 1 25
Grotenfelt's Principles of Modern Dairy Practice. (Wo 11.) nmo, 2 00
Hammarsten's Text-book of Physiological Chemistry. (MandeL) 8vo, 4 00
Helm's Principles of Mathematical Chemistry. (Morgan.) 12x120, 1 50
Bering's Ready Reference Tables (Conversion Factors) i6rr.o morocco, 2 50
Hind's Inorganic Chemistry 8vo, 3 00
* Laboratory Manual for Students nmo, 1 00
Holleman's Text-book of Inorganic Chemistry. (Cooper.) 8vo, 2 50
Text-book of Organic Chemistry. (Walker and Mott.) 8vo, 2 50
* Laboratory Manual of Organic Chemistry. (Walker.) nmo, 1 00
Hopkins's Oil-chemists r Handbook 8vo, 3 00
Jackson's Directions for Laboratory Work in Physiological Chemistry. 8vo, z 25
Keep's Cast Iron . 8vo, 2 50
Ladd's Manual of Quantitative Chemical Analysis nmo, 1 00
Landauer's Spectrum Analysis. (Tingle.) 8vo, 3 00
* Langworthy and Austen. The Occurrence of Aluminium in Vege able
Products, Animal Products, and Natural Waters 8vo, 2 00
Lassar-Cohn's Practical Urinary Analysis. (Lorenz.) i2mo, 1 00
Application of Some General Reactions to Investigations in Organic
Chemistry. (Tingle.) nmo, 1 00
Leach's The Inspection and Analysis of Food with Special Reference to State
Control 8vo, 7 50
Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz. ).i2mo, 1 00
Lodge's Notes on Assaying and Metallurgical Laboratory Experiments. .. .8vo, 3 00
Lunge's Techno-chemical Analysis. (Cohn.) nmo, 1 00
Mandel's Handbook for Bio-chemical Laboratory nmo, 1 50
* Martin's Laboratory Guide to Qualitative Analysis with the Blowpipe . . nmo, 60
Mason's Water-supply. (Considered Principally from a Sanitary Standpoint.)
3d Edition, Rewritten 8vo, 4 00
Examination of Water. (Chemical and Bacteriological) nmo, 1 25
Matthew's The Textile Fibres 8vo, 3 50
Meyer's Determination of Radicles in Carbon Compounds. (Tingle.). . nmo, 1 00
Miller's Manual of Assaying nmo, 1 00
Mixter's Elementary Text-book of Chemistry nmo, 1 50
Morgan's Outline of Theory of Solution and its Results nmo, 1 00
Elements of Physical Chemistry 12 mo, 2 00
Morse's Calculations used in Cane-sugar Factories i6mo, morocco, 1 50
Mulliken's General Method for the Identification of Pure Organic Compounds.
Vol. I Large 8vo, 5 00
O'Brine's Laboratory Guide in Chemical Analysis 8vo, 2 00
O'DriscolTs Notes on the Treatment of Gold Ores 8vo, 2 00
Ostwald'? Conversations on Chemistry. Part One. (Ramsey.) nmo, 1 50
Ostwald's Conversations on Chemistry. Part Two. (Turnbull ). (In Press.)
* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests.
8vo, paper, 50
Pictet's The Alkaloids and their Chemical Constitution. (Biddle.) 8vo, 5 00
Pinner's Introduction to Organic Chemistry. (Austen.) 12 mo, 1 50
Poole's Calorific Power of Fuels 8vo, 3 00
Prescott and Winslow's Elements of Water Bacteriology, with Special Fcfcr-
eace to Sanitary Water Analysis nmo, x 25
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* Reisig's Guide to Piece-dyeing 8vo, 25 00
Richards and Woodman's Air, Water, and Food from a Sanitary Standpoint 8vo, a 00
Richards 's Cost of Living as Modified by Sanitary Science nmo, 1 00
Cost of Food, a Study in Dietaries iamo, x 00
* Richards and Williams's The Dietary Computer 8vo, 1 50
Ricketts and Russell's Skeleton Notes upon Inorganic Chemistry. (Part I.
Non-metallic Elements.) 8vo, morocco, 75
Ricketts and Miller's Notes on Assaying 8to, 3 00
Rideal's Sewage and the Bacterial Purification of Sewage 8vo, 3 50
Disinfection and the Preservation of Food 8vo, 4 00
Rigg's Elementary Manual for the Chemical Laboratory 8to, z 25
Rostoski's Serum Diagnosis. (Bolduan.) iamo, 1 00
Ruddiman's Incompatibilities in Prescriptions 8vo, a 00
Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 3 00
Salkowski's Physiological and Pathological Chemistry. (Orndorff.) 8to, a 50
Schimpf's Text-book of Volumetric Analysis 12 mo, 2 50
Essentials of Volumetric Analysis. nmo, 1 35
Spencer's Handbook for Chemists of Beet-sugar Houses i6mo, morocco. 3 00
Handbook for Sugar Manufacturers and their Chemists. . x6mo, morocco, 2 00
Stockbridge's Rocks and Soils 8vo, 2 50
* Tillman's Elementary Lessons in Heat 8vo, 1 50
* Descriptive General Chemistry 8vo, 3 00
Tread well's Qualitative Analysis. (Hall.) 8vo, 3 00
Quantitative Analysis. (Hall.) 8vo, 4 00
Turneaure and Russell's Public Water-supplies 8vo, 5 00
Van Deventer's Physical Chemistry for Beginners. (Bohwood.) nmo, 1 50
* Walke's Lectures on Explosives 8"o, 4 00
Washington's Manual of the Chemical Analysis of Rocks 8-o, 2 00
Wassermann's Immune Sera : Hemolysins, Cytotoxins, and Precipitins. (Bol-
duan.) nmo, 1 00
Well's Laboratory Guide in Qualitative Chemical Analysis 8vo, 1 50
Short Course in Inorganic Qualitative Chemical Analysis for Engineering
Students nmo, 1 50
Text-book of Chemical Arithmetic nmo, 1 25
Whipple's Microscopy of Drinking-water 8vo, 3 50
Wilson's Cyanide Processes nmo, 1 50
Chlorination Process nmo, 1 50
Wulling's Elementary Course in Inorganic, Pharmaceutical, and Medical
Chemistry nmo, 2 00
CIVIL ENGINEERING.
BRIDGES AND ROOFS. HYDRAULICS. MATERIALS OF ENGINEERING.
RAILWAY ENGINEERING.
Baker's Engineers' Surveying Instruments nmo, 3 00
Bixby's Graphical Computing Table Paper io* X 24* inches. 25
** Burr's Ancient and Modern Engineering and the Isthmian Canal. (Postage,
27 cents additional.) 8vo, 3 50
Comstock's Field Astronomy for Engineers 8vo, 2 50
Davis's Elevation and Stadia Tables 8vo, 1 00
Elliott's Engineering for Land Drainage nmo, 1 50
Practical Farm Drainage nmo, 1 00
♦Fiebeger's Treatise on Civil Engineering 8vo, 5 00
Fohvell's Sewerage. (Designing and Maintenance.) 8vo, 3 00
Freitag's Architectural Engineering. 2d Edition, Rewritten 8vo, 3 50
French and Ives's Stereotomy 8vo, 2 50
Goodhue's Municipal Improvements nmo, x 75
Goodrich's Economic Disposal of Towns' Refuse 8vo, 3 50
Gore's Elements of Geodesy 8vo, 2 50
Hayford's Text-book of Geodetic Astronomy 8vo, 3 00
Hering's Ready Reference Tables (Conversion Factors) i6mo, morocco, a 50
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Howe's Retaining Walk for Earth. xamo, x 23
Johnson's (J. B.) Theory and Practice of Surveying Small 8vo, 4 00
Johnson's (L. J.) Statics by Algebraic and Graphic Methods. 8vo, 2 00
Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.), xamo, a 00
Ifahan's Treatise on Civil Engineering. (1873.) (Wood.) 8vo, 5 00
* Descriptive Geometry 8vo, x 50
Merriman's Elements of Precise Surveying and Geodesy. 8vo, a 50
Elements of Sanitary Engineering 8vo, a 00
Merriman and Brooks's Handbook for Surveyors. i6mo, morocco, a 00
Vugent's Plane Surveying 8vo, 3 So
Ogden's Sewer Design 12 mo, a 00
Patton's Treatise on Civil Engineering 8vo half leather, 7 5®
Heed's Topographical Drawing and Sketching 4to, 5 00
Rideal's Sewage and the Bacterial Purification of Sewat*. 8vo, 3 50
Siebert and Biggin's Modern Stone-cutting and Masonry 8vo, x 50
Smith's Manual of Topographical Drawing. (McMillan.) 8vo, a 50
Sondericker's Graphic Statics, with Applications to Trusses, Beams, and Arches.
8vo, a 00
Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 00
* Trau twine's Civil Engineer's Pocket-book x6mo, morocco, 5 00
Wait's Engineering and Architectural Jurisprudence 8vo, 6 00
Sheep, 6 50
Law of Operations Preliminary to Construction in Engineering and Archi-
tecture 8vo, 5 oo
Sheep, 5 50
Law of Contracts 8vo, 3 00
Warren's Stereotomy — Problems in Stone-cutting 8vo, a 50
Webb's Problems in the Use and Adjustment of Engineering Instruments.
lomo, morocco, z 35
* Wheeler 8 Elementary Course of Civil Engineering 8vo, 4 00
Wilson's Topographic Surveying 8vo, 3 50
BRIDGES AND ROOFS.
Boiler's Practical Treatise on the Construction of Iron Highway Bridges. .8vo, a 00
* Thames River Bridge 4to, paper, 5 00
Burr's Course on the Stresses in Bridges and Roof Trusses, Arched Ribs, and
Suspension Bridges. . . *. 8vo, 3 50
Burr and Falk's Influence Lines for Bridge and Roof Computations . 8vo, 3 00
Du Bole's Mechanics of Engineering. Vol. II Small 4to, 10 00
Foster's Treatise on Wooden Trestle Bridges 4to, 5 00
Fowler's Ordinary Foundations 8vo, 3 50
Greene's Roof Trusses 8vo, 1 25
Bridge Trusses 8vo, 2 50
Arches in Wood, Iron, and Stone 8vo, a 50
Howe's Treatise on Arches 8vo, 4 00
Design of Simple Roof-trusses in Wood and SteeL 8vo, 2 00
Johnson, Bryan, and Turneaure's Theory and Practice in the Designing of
Modern Framed Structures Small 4to, 10 00
Merriman and Jacoby's Text-book on Roofs and Bridges :
Part I. Stresses in Simple Trusses 8vo, 2 50
Part IX Graphic Statics 8vo, 2 50
Part nL Bridge Design 8vo, 2 50
Part IV. Higher Structures 8vo, 2 50
Morison's Memphis Bridge . % 4to, xo 00
Waddell's De Pontibus, a Pocket-book for Bridge Engineers. . i6mo, morocco, 3 00
Specifications for Steel Bridges nmo, 1 25
Wood's Treatise on the Theory of the Construction of Bridges and Roofs . . 8vo, a 00
Wright's Designing of Draw-spans:
Part I. Plate-girder Draws 8vo, a 50
Part II. Riveted- truss and Pin-connected Long-span Draws 8vo, a 50
Two parts in one volume 8vo, 3 50
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HYDRAULICS.
Bazin's Experiments upon the Contraction of the Liquid Vein Issuing from
an Orifice. (Trautwine.) 8vo, 2 00
Bovey*s Treatise on Hydraulics 8vo, 5 00
Church's Mechanics of Engineering 8vo, 6 00
Diagrams of Mean Velocity of Water in Open Channels paper, 1 50
Coffin's Graphical Solution of Hydraulic Problems i6mo, morocco, 2 50
Flather's Dynamometers, and the Measurement of Power 12 mo, 3 00
Folwell's Water-supply Engineering 8vo, 4 00
Frizell's Water-power 8vo, 5 00
Fuertes's Water and Public Health iamo, 1 50
Water-filtration Works 1 amo, 2 50
Ganguillet and Kutter's General Formula for the Uniform Flow of Water in
Rivers and Other Channels. (Hering and Trautwine.) 8vo, 4 00
Hazen's Filtration of Public Water-supply 8vo, 3 00
Hazlehurst's Towers and Tanks for Water-works 8vo, 2 50
Herschel's 115 Experiments on the Carrying Capacity of Large, Riveted, Metal
Conduits 8vo, 2 00
Mason's Water-supply. (Considered Principally from a Sanitary Standpoint. )
8vo, 4 00
Merriman's Treatise on Hydraulics 8vo, 5 00
* Michie's Elements of Analytical Mechanics 8vo, 4 00
Schuyler's Reservoirs for Irrigation, Water-power, and Domestic Water-
supply Large 8vo, 5 00
** Thomas and Watt's Improvement of Rivers. (Post., 44c. additional.) 4to, 6 00
Turneaure and Russell's Public Water-supplies 8vo, 5 00
Wegmann's Design and Construction of Dams 4to, 5 00
Water-supply of the City of New York from 1658 to 1895 4 to, 10 00
Williams and Hazen's Hydraulic Tables 8vo, 1 50
Wilson's Irrigation Engineering Small 8vo, 4 00
Wolff's Windmill as a Prime Mover 8vo, 3 00
Wood's Turbines 8vo, 2 50
Elements of Analytical Mechanics 8vo, 3 00
MATERIALS OF ENGINEERING.
Baker's Treatise on Masonry Construction 8vo, 5 00
Roads and Pavements 8vo, 5 00
Black's United States Public Works Oblong 4to, 5 00
Bovey's Strength of Materials and Theory of Structures 8vo, 7 50
Burr's Elasticity and Resistance of the Materials of Engineering 8vo, 7 50
Byrne's Highway Construction 8vo, 5 00
Inspection of the Materials and Workmanship Employed in Construction.
i6mo, 3 00
Church's Mechanics of Engineering 8vo, 6 00
Du Bois's Mechanics of Engineering. Vol. I Sm*U 4to, 7 50
♦Eckel's Cements, Limes, and Plasters 8vo, 6 00
Johnson's Materials of Construction Large 8vo, 6 00
Fowler's Ordinary Foundations 8vo, 3 50
Keep's Cast Iron 8vo, 2 50
Lanza's Applied Mechanics 8vo, 7 50
Marten's Handbook on Testing Materials. (Henning.) 2 vols 8vo, 7 50
Merrill's Stones for Building and Decoration 8vo, 5 00
Merriman's Mechanics of Materials. 8vo, 5 00
Strength of Materials iamo, 1 00
Metcalf's Steel. A Manual for Steel-users nmo, 2 00
Patton's Practical Treatise on Foundations 8vo, 5 00
Richardson's Modern Asphalt Pavements 8vo, 3 00
Richey's Handbook for Superintendents of Construction i6mo, mor., 4 00
Rockwell's Roads and Pavements in France iamo, 1 35
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Sabin's Industrial and Artistic Technology of Paints and Varnish. 8vo,
Smith's Materials of Machines nmo,
Snow's Principal Species of Wood 8vo,
Spalding's Hydraulic Cement nmo,
Text-book on Roads and Pavements iamo,
Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo,
Thurston's Materials of Engineering. 3 Parts 8vo,
Part I. Non-metallic Materials of Engineering and Metallurgy 8vo,
Part II. Iron and Steel. 8vo,
Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo,
Thurston's Text-book of the Materials of Construction 8vo,
Tillson's Street Pavements and Paving Materials 8vo,
Waddell's De Pontibus. ( * Pocket-book for Bridge Engineers.). . x6mo, mor.,
Specifications for Stt . i Bridges 12 mo,
Wood's (De V.) Treatise on the Resistance of Materials, and an Appendix on
the Preservation of Timber 8vo,
Wood's (De V.) Elements of Analytical Mechanics 8vo,
Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and
Steel 8vo, 4 00
RAILWAY ENGINEERING.
Andrew's Handbook for Street Railway Engineers 3x5 inches, morocco, z 25
Berg's Buildings and Structures of American Railroads 4 to, 5 00
Brook's Handbook of Street Railroad Location. i6mo, morocco, x 50
Butt's Civil Engineer's Field-book. x6mo, morocco, 2 50
Crandall's Transition Curve i6mo, morocco, 1 50
Railway and Other Earthwork Tables. 8vo, 1 50
Dawson's "Engineering" and Electric Traction Pocket-book. . i6mo, morocco, 5 00
Dredge's History of the Pennsylvania Railroad: (1879) '. . .Paper, 5 00
* Drinker's Tunnelling, Explosive Compounds, and Rock Drills. 4to, half mor., 25 00
Fisher's Table of Cubic Yards Cardboard, - 25
Godwin's Railroad Engineers' Field-book and Explorers' Guide. . . i6mo, mor., 2 50
Howard's Transition Curve Field-book 16 mo, morocco, 1 50
Hudson's Tables for Calculating the Cubic Contents of Excavations and Em-
bankments 8vo, x 00
Molitor and Beard'B Manual for Resident Engineers i6mo, 1 00
Nagle's Field Manual for Railroad Engineers i6mo, morocco, 3 00
Philbrick's Field Manual for Engineers i6mo, morocco, 3 00
Searles's Field Engineering i6mo, morocco, 3 00
Railroad Spiral. i6mo, morocco, 1 50
Taylor's Prismoidal Formulae and Earthwork 8vo, 1 50
* Trautwine's Method of Calculating the Cube Contents of Excavations and
Embankments by the Aid of Diagrams 8vo, 2 00
The Field Practice of Laying Out Circular Curves for Railroads.
i2ino, morocco,
Cross-section Sheet Paper,
Webb's Railroad Construction x6mo, morocco,
Wellington's Economic Theory of the Location of Railways Small 8vo,
DRAWING.
Barr's Kinematics of Machinery 8vo,
* Bartlett's Mechanical Drawing 8vo,
* " " Abridged Ed 8vo,
Coolidge's Manual of Drawing 8vo, paper
Coolidge and Freeman's Elements of General Drafting for Mechanical Engi-
neers Oblong 4to,
Durley's Kinematics of Machines 8vo,
Bmch's Introduction to Projective Geometry and its Applications 8vo.
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Hill's Text-book on Shades and Shadows, and Perspective 8vo, 2 00
Jamison's Elements of Mechanical Drawing 8vo, 2 50
Advanced Mechanical Drawing 8vo, 2 00
Jones's Machine Design:
Part I. Kinematics of Machinery. 8vo, 1 50
Part II. Form, Strength, and Proportions of Parts 8vo, 3 00
MacCord's Elements of Descriptive Geometry 8vo, 3 00
Kinematics; or. Practical Mechanism. 8vo, 5 00
Mechanical Drawing. 4to, 4 00
Velocity Diagrams 8vo, 1 50
* Mahan's Descriptive Geometry and Stone-cutting 8vo, x 50
Industrial Drawing. (Thompson.) 8vo, 3 50
Moyer's Descriptive Geometry. 8vo, 2 00
Reed's Topographical Drawing and Sketching 4to, 5 00
Reid's Course in Mechanical Drawing 8vo, 2 00
Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 3 00
Robinson's Principles of Mechanism. 8vo, 3 00
Schwamb and Merrill's Elements of Mechanism 8vo, 3 00
Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 2 50
Warren's Elements of Plane and Solid Free-hand Geometrical Drawing. 1 a mo, x 00
Drafting Instruments and Operations xamo. x 25
Manual of Elementary Projection Drawing iamo, 1 se
Manual of Elementary Problems in the Linear Perspective of Form and
Shadow 1 2 mo, 1 00
Plane Problems in Elementary Geometry nmo, 1 25
Primary Geometry nmo, 75
Elements of Descriptive Geometry, Shadows, and Perspective 8vo, 3 50
General Problems of Shades and Shadows 8vo, 3 00
Elements of Machine Construction and Drawing 8vo, 7 50
Problems, Theorems, and Examples in Descriptive Geometry 8vo, 2 50
Weisbach's Kinematics and Power of Transmission. (Hermann and Klein )8vo, 5 00
Whelpley's Practical Instruction in the Axt of Letter Engraving nmo, 2 00
Wilson's (H. M.) Topographic Surveying 8vo, 3 50
Wilson's (V. T.) Free-hand Perspective 8vo, 2 50
Wilson's (V. T.) Free-hand Lettering 8vo, 1 00
Woolf's Elementary Course in Descriptive Geometry Large 8vo, 3 00
ELECTRICITY AND PHYSICS.
Anthony and Brackett's Text-book of Physics. (Magie.) Small 8vo, 3 00
Anthony's Lecture-notes on the Theory of Electrical Measurements. . . . 12 mo, 1 00
Benjamin's History of Electricity 8vo, 3 00
Voltaic CelL 8vo, 3 00
Classen's Quantitative Chemical Analysis by Electrolysis. ( Bo It wood.). 8 vo, 3 00
Crehore and Squier's Polarizing Photo-chronograph 8vo, 3 00
Dawson's "Engineering" and Electric Traction Pocket-book. i6mo. morocco, 5 00
Dolezalek's Theory of the Lead Accumulator (Storage Battery). (Von
Ende.) 12 mo, 2 50
Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 00
Flather's Dynamometers, and the Measurement of Power nmo, 3 00
Gilbert's De Magnete. (Mottelay.) 8vo, 2 50
Hanchett's Alternating Currents Explained nmo, 1 00
Bering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 50
Holman's Precision of Measurements 8vo. 2 00
Telescopic Mirror-scale Method, Adjustments, and Tests .... Large 8vo, 75
Kinzbrunner's Testing of Continuous-Current Machines 8vo. 2 00
Landauer's Spectrum Analysis. (Tingle. ) 8vo. 3 00
Le Chateliens High-temperature Measurements. (Boudouard — Burgess.) iamo. 3 00
Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz. ) iamo, 1 00
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* Lyons's Treatise en Electromagnetic Phenomena. Volt. I. and IL 8yo, each, 6 oo
* Michie's Elements of Ware Motion Relating to Sound and Light 8to, 4 00
Hiaudef t Elementary Treatise on Electric Batteries. (Fishback.) 12 mo, 2 so
* Rosenberg's Electrical Engineering. (He Wane Gee — Kinzbrunner.). . 8vo, x 50
Ryan, Norm, and Hoxie's Electrical Machinery. Vol I. 8vo, 2 50
Thurston's Stationary Steam-engines 8to, 2 50
* Tillman's Elementary Lessons in Heat 8vo, x 50
Tory and Pitcher's Manual of Laboratory Physics Small 8vo, 2 00
Ulke's Modern Electrolytic Copper Refining 8vo, 3 00
LAW.
* Davis's Elements of Law 8vo, 2 50
* Treatise on the Military Law of United States. 8vo, 7 00
* Sheep, 7 50
Manual for Courts-martial. i6mo, morocco, x 50
Wait's Engineering and Architectural Jurisprudence 8vo, 6 00
Sheep, 6 50
Law of Operations Preliminary to Construction in Engineering and Archi-
tecture 8vo, 5 00
Sheep, 5 50
Law of Contracts. 8vo, 3 00
Winthrop's Abridgment of Military Law 12010, 2 50
MANUFACTURES.
Bernadou's Smokeless Powder— Nitro-cellulose and Theory of the Cellulose
Molecule i2mo, 2 50
Bolland's Iron Founder umo, 2 50
*' The Iron Founder," Supplement i2mo, 2 50
Encyclopedia of Founding and Dictionary of Foundry Terms Used in the
Practice of Moulding 1 2mo, 3 00
Eissler's Modern High Explosives 8vo, 4 00
Effront's Enzymes and their Applications. (Prescott.) 8vo, 3 00
Fitzgerald's Boston Machinist 12010, 1 00
Ford's Boiler Making for Boiler Makers i8mo, 1 00
Hopkin's Oil-chemists' Handbook 8vo, 3 00
Keep's Cast Iron 8vo, 2 50
Leach's The Inspection and Analysis of Food with Special Reference to State
Control. Large 8vo, 7 50
Matthews's The Textile Fibres 8vo, 3 50
Metcalf's Steel. A Manual for Steel-users nmo, 2 00
Metcalfe's Cost of Manufactures — And the Administration of Workshops. 8 vo, 5 00
Meyer's Modern Locomotive Construction 4to, 10 00
Morse's Calculations used in Cane-sugar Factories i6rao, morocco, 1 50
* Reisig's Guide to Piece-dyeing 8vo, 25 00
Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 3 00
Smith's Press-working of Metals 8vo, 3 00
Spalding's Hydraulic Cement nmo, 2 00
Spencer's Handbook for Chemists of Beet-sugar Houses. . . . i6mo, morocco, 3 00
Handbook for Sugar Manufacturers and their Chemists . i6mo, morocco, 2 00
Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 00
Thurston's Manual of Steam-boilers, their Designs, Construction and Opera-
tion 8vo, 5 00
* Walke's Lectures on Explosives 8vo, 4 00
Ware's Manufacture of Sugar. (In press.)
West's American Foundry Practice nmo, 2 50
Moulder's Text-book. nmo, 2 50
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Wolffs Windmill as a Prime Mover 8vo, 3 o*
Wood's Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. .8vo, 4 00
MATHEMATICS.
Baker's Elliptic Functions 8to, i 5«
* Bass's Elements of Differential Calculus 12 mo, 4 00
Briggs's Elements of Plane Analytic Geometry 12 mo, 1 00
Compton's Manual of Logarithmic Computations. nmo, 1 50
Davis's Introduction to the Logic of Algebra 8vo, 1 50
* Dickson's College Algebra Large iamo, 1 50
* Introduction to the Theory of Algebraic Equations Large 12 mo, x 25
Emch's Introduction to Projective Geometry and its Applications 8vo, 2 50
Halsted's Elements of Geometry 8vo, 1 75
Elementary Synthetic Geometry. 8vo, x 50
Rational Geometry 12 mo, 1 75
* Johnson's (J. B.) Three-place Logarithmic Tables: Vest-pocket size. paper, 15
xoo copies for 5 00
* Mounted on heavy cardboard, 8:< 10 inches, 25
xo copies for 2 00
Johnson's (W. W.) Elementary Treatise on Differential Calculus. .Small 8vo, 3 00
Johnson's (W. W.) Elementary Treatise on the Integral Calculus. Small 8vo, 1 50
Johnson's (W. W.) Curve Tracing in Cartesian Co-ordinates. 12 mo, 1 00
Johnson's (W. W.) Treatise on Ordinary and Partial Differential Equations.
Small 8vo, 3 50
Johnson's (W. W.) Theory of Errors and the Method of Least Squares. i2mo, 1 50
* Johnson's (W. W.) Theoretical Mechanics nao, 3 00
Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) . ismo, 2 00
* Ludlow and Bass. Elements of Trigonometry and Logarithmic and Other
Tables 8vo, 3 00
Trigonometry and Tables published separately Each, 2 00
* Ludlow's Logarithmic and Trigonometric Tables 8vo, x 00
Maurer'B Technical Mechanics 8vo, 4 00
Merriman and Woodward's Higher Mathematics. 8vo, 5 00
Merriman's Method of Least Squares 8vo, 2 oe
Rice and Johnson's Elementary Treatise on the Differential Calculus. . Sm. 8vo, 3 00
Differential and Integral Calculus. 2 vols, in one Small 8vo, 2 50
Wood's Elements of Co-ordinate Geometry 8vo, 2 00
Trigonometry: Analytical, Plane, and Spherical 12 mo, x 00
MECHANICAL ENGINEERING.
MATERIALS OF ENGINEERING, STEAM-ENGINES AND BOILERS.
Bacon's Forge Practice 12010, 1 50
Baldwin's Steam Heating for Buildings 12 mo, 2 50
Barr's Kinematics of Machinery 8vo, 2 50
* Bartlett's Mechanical Drawing 8vo, 3 00
♦ " " " Abridged Ed 8vo, 150
Benjamin's Wrinkles and Recipes iamo, 2 00
Carpenter's Experimental Engineering 8vo, 6 00
Heating and Ventilating Buildings 8vo, 4 00
Cary's Smoke Suppression in Plants using Bituminous CoaL (In Prepara-
tion.)
Clerk's Gas and Oil Engine Small 8vo, 4 00
Coolidge's Manual of Drawing 8vo, paper, 1 00
Coolidge and Freeman's Elements of General Drafting for Mechanical En-
gineers Oblong 4to, 2 50
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Cromwell's Treatise on Toothed Gearing iamo, i 50
Treatise on Belts and Pulleys. 12 mo, x 50-
Durley's Kinematics of Machines. 8yo, 4 00
Flather's Dynamometers and the Measurement of Power. 12 mo, 3 00
Rope Driving 12 mo, 2 00
Gill's Gas and Fuel Analysis for Engineers iamo, x 25
Hall's Car Lubrication. iamo, 1 00
Bering's Ready Reference Tables (Conversion Factors) x6mo, morocco, 2 so
Hutton's The Gas Engine 8vo, 5 00
Jamison's Mechanical Drawing 8vo, 2 5a
Jones's Machine Design:
Part L Kinematics of Machinery. 8vo, x 50-
Part H. Form, Strength, and Proportions of Parts 8vo, 3 00
Kent's Mechanical Engineers' Pocket-book. x6mo, morocco, 5 00
Kerr's Power and Power Transmission. 8vo, 2 00
Leonard's Machine Shop, Tools, and Methods 8vo, 4 00
•Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dean.) . . 8vo, 4 00
MacCord's Kinematics; or, Practical Mechanism. 8vo, 5 00
Mechanical Drawing 4to, 4 00
Velocity Diagrams 8vo, x 50
Mahan's Industrial Drawing. (Thompson.) 8vo, 3 50
Poole s Calorific Power of Fuels 8vo, 3 00
Reid's Course in Mechanical Drawing 8vo, 2 00
Text-book of Mechanical Drawing and Elementary Machine Design. 8 vo, 3 oe
Richard's Compressed Air 12010, 1 50
Robinson's Principles of Mechanism 8vo, 3 00
Schwamb and Merrill's Elements of Mechanism. 8vo, 3 00
Smith's Press-working of Metals. 8vo, 3 00
Thurston's Treatise on Friction and Lost Work in Machinery and Mill
Work 8vo, 3 00
Animal as a Machine and Prime Motor, and the Laws of Energetics . 12 mo, x 00
Warren's Elements of Machine Construction and Drawing 8vo, 7 50
Weisbach's Kinematics and the Power of Transmission. (Herrmann —
Klein.) 8vo, 5 00
Machinery of Transmission and Governors. (Herrmann — Klein.). .8vo, 5 00
Wolff's Windmill as a Prime Mover 8vo, 3 00
Wood's Turbines 8vo, 2 50-
MATERIALS OF ENGINEERING.
Bovey*B Strength of Materials and Theory of Structures 8vo, 7 50
Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edition.
Reset. 8vo, 7 50
Church's Mechanics of Engineering 8vo, 6 00
Johnson's Materials of Construction. 8vo, 6 00
Keep's Cast Iron. 8vo, 2 5a
Lanza's Applied Mechanics. 8vo, 7 50
Martens's Handbook on Testing Materials. (Henning.) 8vo, 7 50
Merriman's Mechanics of Materials. 8vo, 5 00
Strength of Materials iamo, x 00
Metcalf* s SteeL A manual for Steel-users iamo. 2 00
Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 3 00
Smith's Materials of Machines iamo, x 00
Thurston's Materials of Engineering 3 vols., 8vo, 8 00
Part II. Iron and SteeL 8vo, 3 50
Part in. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo, 2 5*
Text-book of the Materials of Construction. 8vo, 5 00,
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Wood's (De V.) Treatise on the Resistance of Materials and an Appendix on
the Preservation of Timber 8vo, 2 oO
Wood's (De V.) Elements of Analytical Mechanics. 8vo, 3 00
Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and
Steel 8vo f 4 00
STEAM-ENGINES AND BOILERS.
Berry's Temperature-entropy Diagram. i2mo, x 25
Carnot's Reflections on the Motive Power of Heat. (Thurston.) xamo, 1 50
Dawson's "Engineering" and Electric Traction Pocket-book. . . . i6mo, mor., 5 00
Ford's Boiler Making for Boiler Makers i8mo, 1 00
Goss's Locomotive Sparks 8vo, 2 00
Hemcnway's Indicator Practice and Steam-engine Economy iamo, 2 00
Hutt on's Mechanical Engineering of Power Plants 8vo, 5 00
Heat and Heat-engines. 8vo, 5 00
Kent's Steam boiler Economy 8vo, 4 00
Kneass's Practice and Theory of the Injector 8vo, 1 50
MacCord's Slide-valves. 8vo, 2 00
Meyer's Modern Locomotive Construction 4to, zo 00
Peabody's Manual of the Steam-engine Indicator iamo. 1 50
Tables of the Properties of Saturated Steam and Other Vapors 8vo, x 00
Thermodynamics of the Steam-engine and Other Heat-engines 8vo, 5 00
Valve-gears for Steam-engines. 8vo, 2 50
Peabody and Miller's Steam-boilers 8vo, 4 00
Pray's Twenty Years with the Indicator Large 8vo, 2 50
Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors.
(Osterberg.) iamo, 1 25
Reagan's Locomotives: Simple Compound, and Electric i2mo, 2 50
Rontgen's Principles of Thermodynamics. (Du Bois.) 8vo, 5 00
Sinclair's Locomotive Engine Running and Management 12 mo, 2 00
Smart's Handbook of Engineering Laboratory Practice 12 mo, 2 50
Snow's Steam-boiler Practice 8vo, 3 00
Spangler's Valve-gears 8vo, 2 50
Notes on Thermodynamics i2mo, 1 00
Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 3 00
Thurston's Handy Tables 8vo, 1 50
Manual of the Steam-engine 2 vols., 8vo, 10 00
Part I. History, Structure, and Theory 8vo, 6 00
Part IL Design, Construction, and Operation 8vo, 6 00
Handbook of Engine and Boiler Trials, and the Use of the Indicator and
the Prony Brake 8vo, 5 00
Stationary Steam-engines. 8vo, 2 50
Steam-boiler Explosions in Theory and in Practice iamo, 1 50
Manual of Steam-boilers, their Designs, Construction, and Operation 8vo, 5 00
Weis bach's Heat, Steam, and Steam-engines. (Du Bois.) 8vo, 5 00
Whitham's Steam-engine Design 8vo, 5 00
Wilson's Treatise on Steam-boilers. (Flather.) i6mo, 2 50
Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .8vo, 4 00
MECHANICS AND MACHINERY.
Ban's Kinematics of Machinery 8vo, 2 50
Bovey's Strength of Materials and Theory of Structures 8vo, 7 50
Chase's The Art of Pattern-making *. . . . iamo, 2 50
C.hurch!s Mechanics of Engineering 8vo, 6 00
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Church's Notes and Examples in Mechanics. 8vo, a oo
Compton's First Lessons in Metal-working. xamo, x 50
Compton and De Oroodt's The Speed Lathe 12 mo, x 50
Cromwell's Treatise on Toothed Gearing xamo, x 50
Treatise on Belts and Pulleys. iamo, x 50
Dana's Text-book of Elementary Mechanics for Colleges and Schools. . 12 mo, x 50
Dingey's Machinery Pattern Making 12 mo, 2 00
Dredge's Record of the Transportation Exhibits Building of the World's
Columbian Exposition of 1893 4to half morocco, 5 00
Du Bois's Elementary Principles of Mechanics:
Vol L Kinematics..* 8vo, 350
VoL II. Statics. 8vo, 4 00
VoL m. Kinetics. 8vo, 3 50
Mechanics of Engineering. VoL I Small 4to, 7 50
VoL H. Small 4to, 10 00
Durley's Kinematics of Machines. 8vo, 4 00
Fitzgerald's Boston Machinist i6mo, x 00
Flather's Dynamometers, and the Measurement of Power 12 mo, 3 00
Rope Driving xamo, 2 00
Goss's Locomotive Sparks. 8vo, 2 00
Hall's Car Lubrication 12 mo, x 00
Holly's Art of Saw Filing i8mo, 75
James's Kinematics of a Point and the Rational Mechanics of a Particle. Sm.8vo,2 00
* Johnson's (W. W.) Theoretical Mechanics iamo, 3 00
Johnson's (L. J.) Statics by Graphic and Algebraic Methods 8vo, 2 00
Jones's Machine Design :
Part L Kinematics of Machinery 8vo,
Part II. Form, Strength, and Proportions of Parts. . . . , 8vc,
Kerr's Power and Power Transmission 8vo,
Lanza's Applied Mechanics 8vo,
Leonard's Machine Shop, Tools, and Methods 8vo,
♦Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dean. ) 8vo,
MacCord's Kinematics; or, Practical Mechanism 8vo,
Velocity Diagrams 8vo,
Maurer's Technical Mechanics 8vo,
Merriman's Mechanics of Materials 8vo,
* Elements of Mechanics iamo,
* Michie's Elements of Analytical Mechanics 8vo,
Reagan's Locomotives: Simple, Compound, and Electric umo,
Reid's Course in Mechanical Drawing 8vo.
Text-book of Mechanical Drawing and Elementary Machine Design. 8 vo,
Richards's Compressed Air i2mo,
Robinson's Principles of Mechanism. 8vo,
Ryan, Norris, and Hoxie's Electrical Machinery. VoL 1 8vo,
Schwamb and Merrill's Elements of Mechanism 8vo,
Sinclair's Locomotive-engine Running and Management X2mo,
Smith's (O.) Press-working of Metals 8vo,
Smith's (A. W.) Materials of Machines xamo,
Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo,
Thurston's Treatise on Friction and Lost Y/ork in Machinery and Mill
Work 8vo,
Animal as a Machine and Prime Motor, and the Laws of Energetics.
i2mo,
Warren's Elements of Machine Construction and Drawing 8vo,
Weisbach's Kinematics and Power of Transmission. (Herrmann — Klein. ) . 8vo,
Machinery of Transmission and Governors. (Herrmann — Klein. ).8vo,
Wood's Elements of Analytical Mechanics 8vo.
Principles of Elementary Mechanics 12 mo,
Turbines • 8vo.
The World's Columbian Exposition of 1893 4 to,
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METALLURGY.
Egleston's Metallurgy of Silver, Gold, and Mercury:
Vol. t Silver 8vo, 7 50
VoL DL Gold and Mercury 8vo, 750
** Ilea's Lead-smelting. (Postage 9 cents additional) x2mo, 2 50
Keep's Cast Iron 8vo, 2 50
Kunhardt's Practice of Ore Dressing in Europe 8vo, x 50
Le Chatelier's High-temperature Measurements. (Boudouard — Burgess. )i2mo, 3 00
MetcalTs Steel A Manual for Steel-users iamo, 2 00
Smith's Materials of Machines. X2mo, x 00
Thurston's Materials of Engineering. In Three Parts. 8vo 8 00
Part IX Iron and SteeL 8vo. 3 50
Part in. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents. .' 8vo, 2 50
Ulke's Modern Electrolytic Copper Refining 8vo, 3 00
MINERALOGY.
Barringer's Description of Minerals of Commercial Value. Oblong, morocco, 2 50
Boyd's Resources of Southwest Virginia 8vo, 3 00
Map of Southwest Virignia. Pocket-book form. 2 00
Brush's Manual of Determinative Mineralogy. (Penneld.) 8vo, 4 00
Chester's Catalogue of Minerals 8vo, paper, 1 00
Cloth, x 25
Dictionary of the Names of Minerals 8vo, 3 50
Dana's System of Mineralogy Large 8vo, half leather, 12 50
First Appendix to Dana's New " System of Mineralogy." Large 8vo, x 00
Text-book of Mineralogy 8vo, 4 00
Minerals and How to Study Them X2mo, x 50
Catalogue of American Localities of Minerals Large 8vo, 1 00
Manual of Mineralogy and Petrography iamo . 2 00
Douglas's Untechnical Addresses on Technical Subjects 12 mo, 1 00
Eakle's Mineral Tables 8vo, x 25
Egleston's Catalogue of Minerals and Synonyms 8vo, 2 50
Hussak's The Determination of Rock-forming Minerals. ( Smith.). Small 8vo. 2 00
Merrill's Non-metallic Minerals: Their Occurrence and Uses 8vo, 4 00
* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests.
8vo paper, o 50
Rosenbusch's Microscopical Physiography of the Rock-making Minerals
(Iddings.) 8vo. 5 00
* Tillman's Text-book of Important Minerals and Rocks 8vo. 2 00
Williams's Manual of Lithology 8vo, 3 00
MINING.
Beard's Ventilation of Mines 12 mo. 2 50
Boyd's Resources of Southwest Virginia 8vo. 3 00
Map of Southwest Virginia Pocket book form,
Douglas's Untechnical Addresses on Technical Subjects iamo.
* Drinker's Tunneling, Explosive Compounds, and Rock Drills 4t0.tif mor
Eissler's Modern High Explosives 8vo
Fowler's Sewage Works Analyses i2mo.
Goodyear's Coal-mines of the Western Coast of the United States iamo.
Ihlseng's Manual of Mining . .8va.
** Iles's Lead-smelting. (Postage oc. additional.) - iamo.
Kunhardt's Practice of Ore Dressing in Europe .8vo,
O'DriscolTs Notes on the Treatment of Gold Ores 8vo.
* Walke's Lectures on Explosives. 8vo,
Wilson's Cyanide Processes. iamo,
Chlorination Process. , . .i2mo,
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Wilson's Hydraulic and Placer Mining 12 mo,
Treatise on Practical and Theoretical Mine Ventilation iamo,
SANITARY SCIENCE.
Bashore's Sanitation of a Country House xamo,
FolwelTs Sewerage. (Designing, Construction, and Maintenance.) 8vo,
Water-supply Engineering 8to,
Fuertes's Water and Public Health. 12 mo,
Water-filtration Works iamo,
Gerhard's Guide to Sanitary House-inspection i6mo,
Goodrich's Economic Disposal of Town's Refuse Demy 8vo,
Hazen's Filtration of Public Water-supplies 8vo,
Leach's The Inspection and Analysis of Food with Special Reference to State
Control 8vo,
Mason's Water-supply. (Considered principally from a Sanitary Standpoint) 8vo,
Examination of Water. (Chemical and Bacteriological.) i2mo,
Me rri man's Elements of Sanitary Engineering 8ro,
Ogden's Sewer Design 12 mo,
Prescott and Winslow's Elements of Water Bacteriology, with Special Refer-
ence to Sanitary Water Analysis iamo t
* Price's Handbook on Sanitation nmo,
Richards's Cost of Food. A Study in Dietaries , nmo,
Cost of Living as Modified by Sanitaiy Science nmo,
Richards and Woodman's Air, Water, and Food from a Sanitary Stand-
point 8vo,
* Richards and Williams's The Dietary Computer 8vo,
Rideal's Sewage and Bacterial Purification of Sewage 8vo,
Turneaure and Russell's Public Water-supplies 8vo,
Von Behring's Suppression of Tuberculosis. (Bolduan.) nmo,
Whipple's Microscopy of Drinking-water 8vo,
Woodhull's Notes on Military Hygiene x6mo,
MISCELLANEOUS.
De Fursac's Manual of Psychiatry. (Rosanoff and Collins.). . . .Large nmo, 2 50
Emmons's Geological Guide-book of the Rocky Mountain Excursion of the
International Congress of Geologists Large 8vo, x 50
Ferrel's Popular Treatise on the Winds 8vo. 4 00
Haines's American Railway Management nmo, 2 50
Mott's Composition, Digestibility, and Nutritive Value of Food. Mounted chart, x 25
Fallacy of the Present Theory of Sound i6mo, 1 00
Ricketts's History of Rensselaer Polytechnic Institute, 1 824-1 894. .Small 8vo, 3 00
Rostoski's Serum Diagnosis. (Bolduan.) nmo, 1 00
Rotherham's Emphasized New Testament Large 8vo, 2 00
Steel's Treatise on the Diseases of the Dog 8vo, 3 50
Totten's Important Question in Metrology 8vo, 2 50
The World's Columbian Exposition of 1893 4*0, 1 00
Von Behring's Suppression of Tuberculosis. (Bolduan.) nmo, 1 00
Winslow's Elements of Applied Microscopy nmo, 1 50
Worcester and Atkinson. Small Hospitals, Establishment and Maintenance;
Suggestions for Hospital Architecture : Plans for Small Hospital nmo, x 25
HEBREW AND CHALDEE TEXT-BOOKS.
Green's Elementary Hebrew Grammar X2mo, 1 25
Hebrew Chrestomathy 8vo, a 00
Gesenius's Hebrew and Chaldee Lexicon tf the Old Testament Scriptures.
(Tregelles.) Small 4to, half morocco, 5 00
Lettems's Hebrew Bible 8 ™» 2 2g
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